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a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
\(a.\frac{7x-3}{x-1}=\frac{2}{3}\\\Leftrightarrow \frac{3\left(7x-3\right)}{3\left(x-1\right)}= \frac{2\left(x-1\right)}{3\left(x-1\right)}\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\\Leftrightarrow 3\left(7x-3\right)-2\left(x-1\right)=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\frac{7}{19}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\\ \Leftrightarrow\frac{4\left(3-7x\right)}{2\left(1+x\right)}=\frac{1\left(1+x\right)}{2\left(1+x\right)}\\\Leftrightarrow 4\left(3-7x\right)=1\left(1+x\right)\\ \Leftrightarrow4\left(3-7x\right)-1\left(1+x\right)=0\\ \Leftrightarrow12-28x-1-x=0\\ \Leftrightarrow11-29x=0\\ \Leftrightarrow-29x=-11\\ \Leftrightarrow x=\frac{-11}{-29}=\frac{11}{29}\)
\(c.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\\ \Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x+2\right)\left(3x-1\right)}\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)-\left(5x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow15x^2-5x-3x+1-15x^2-10x+21x+14=0\\ \Leftrightarrow3x+15=0\\\Leftrightarrow 3x=-15\\\Leftrightarrow x=-5\)
\(d.\frac{4x+7}{x-1}=\frac{12x+5}{3x+4}\\\Leftrightarrow \frac{\left(4x+7\right)\left(3x+4\right)}{\left(x-1\right)\left(3x+4\right)}=\frac{\left(12x+5\right)\left(x-1\right)}{\left(3x+4\right)\left(x-1\right)}\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow \left(4x+7\right)\left(3x+4\right)-\left(12x+5\right)\left(x-1\right)=0\\ \Leftrightarrow12x^2+16x+21x+28-12x^2-12x+5x-5=0\\ \Leftrightarrow30x+23=0\\ \Leftrightarrow30x=-23\\ \Leftrightarrow x=\frac{-23}{30}\)
\(e.\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\\ \Leftrightarrow1+3\left(x-2\right)=3-x\\\Leftrightarrow 1+3x-6=3-x\\\Leftrightarrow 1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\)
\(f.\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8\left(x-7\right)}{x-7}=\frac{1}{x-7}\\ \Leftrightarrow8-x-8\left(x-7\right)=1\\ \Leftrightarrow8-x-8\left(x-7\right)-1=0\\\Leftrightarrow 8-x-8x+56-1=0\\\Leftrightarrow 63-9x=0\\\Leftrightarrow -9x=-63\\ \Leftrightarrow x=\frac{-63}{-9}=7\)
\(g.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\\Leftrightarrow \frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\\\Leftrightarrow \left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)-20=0\\ \Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\)
\(j.\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\\\Leftrightarrow \frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{2.2x}{2\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\\\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)-4x=0\\\Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right. \)
a/ \(7x-5=13-5x\)
\(\Leftrightarrow7x+5x=13+5\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\frac{3}{2}\)
b/\(5\left(2x-3\right)-4\left(5x-7\right)=19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28=19-2x-22\)
\(\Leftrightarrow10x-20x+2x=19-22-28+15\)
\(\Leftrightarrow-8x=-16\)
\(\Leftrightarrow x=2\)
c/ \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\frac{7\left(2x-1\right)-3\left(5x+2\right)-21\left(x+13\right)}{21}=0\)
\(\Leftrightarrow14x-7-15x-6-21x-273=0\)
\(\Leftrightarrow-22x-286=0\)
\(\Leftrightarrow x=-13\)
e/ \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)\left(x+2\right)-\left(x+1\right)\left(x+2\right)-\left(3x-11\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x^2-4\right)-\left(x^2+3x+2\right)-\left(3x^2-17x+22\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow2x^2-8-x^2-3x-2-3x^2+17x-22=0\)
\(\Leftrightarrow-2x^2+14x-32=0\)
\(\Leftrightarrow x^2-7x+16=0\)
\(\Leftrightarrow x=\frac{-\left(-7\right)\pm\sqrt{\left(-7\right)^2-4.1.16}}{2}\)
\(\Leftrightarrow x=\frac{7\pm\sqrt{-15}}{2}\left(ktm\right)\)
\(\Leftrightarrow x\in\varnothing\)
Bài 1:
a) \(7x-5=13-5x\)
\(\Leftrightarrow7x+5x=13+5\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=18:12\)
\(\Leftrightarrow x=\frac{3}{2}.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{3}{2}\right\}.\)
b) \(5.\left(2x-3\right)-4.\left(5x-7\right)=19-2.\left(x+11\right)\)
\(\Leftrightarrow10x-15-\left(20x-28\right)=19-\left(2x+22\right)\)
\(\Leftrightarrow10x-15-20x+28=19-2x-22\)
\(\Leftrightarrow13-10x=-3-2x\)
\(\Leftrightarrow13+3=-2x+10x\)
\(\Leftrightarrow16=8x\)
\(\Leftrightarrow x=16:8\)
\(\Leftrightarrow x=2.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2\right\}.\)
c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
\(\Leftrightarrow\frac{7.\left(2x-1\right)}{7.3}-\frac{3.\left(5x+2\right)}{3.7}=\frac{21.\left(x+13\right)}{21}\)
\(\Leftrightarrow\frac{14x-7}{21}-\frac{15x+6}{21}=\frac{21x+273}{21}\)
\(\Leftrightarrow14x-7-\left(15x+6\right)=21x+273\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow-x-13=21x+273\)
\(\Leftrightarrow-x-21x=273+13\)
\(\Leftrightarrow-22x=286\)
\(\Leftrightarrow x=286:\left(-22\right)\)
\(\Leftrightarrow x=-13.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-13\right\}.\)
Chúc bạn học tốt!
2. \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\) (ĐKXĐ:\(x\ne1,x\ne2\))
\(\Leftrightarrow\frac{1}{x-1}+\frac{7}{2-x}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Leftrightarrow\frac{2-x+7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Rightarrow2-x+7\left(x-1\right)=1\)
\(\Leftrightarrow2-x+7x-7=1\)
\(\Leftrightarrow-x+7x=1-2+7\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
Vậy phương trình trên vô nghiệm
ko phan tich duoc nha bn
chuc bn hoc gioi
happy new year
\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
<=> \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
<=> \(\frac{3\left(2x-1\right)}{5\cdot3}-\frac{5\left(x-2\right)}{3\cdot5}-\frac{x+7}{15}=0\)
<=> \(\frac{6x-3-5x+10-x-7}{15}=0\)
<=> \(\frac{-14}{15}=0\)
=> PT vô nghiệm
a)
\(\frac{7}{x-5}-2=\frac{3}{5-x}\\ \Leftrightarrow\frac{-7}{5-x}-2-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7}{5-x}-\frac{10-2x}{5-x}-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7-10+2x-3}{5-x}=0\\ \Leftrightarrow\frac{2x-20}{5-x}=0\\ \Rightarrow2x-20=0\\ \Rightarrow x=10\)
b)
\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}\\ \Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4-x-1-3x+11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{6-2x}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Rightarrow6-2x=0\\ \Rightarrow x=3\)
c)
\(\frac{1}{x}-\frac{x+2}{x-2}=\frac{2}{x\cdot\left(2-x\right)}\\ \Leftrightarrow\frac{1}{x}-\frac{x-2}{2-x}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x}{x\cdot\left(2-x\right)}-\frac{x^2-2x}{x\cdot\left(2-x\right)}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x-x^2+2x-2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{x-x^2}{x\cdot\left(2-x\right)}=0\\ \Rightarrow x-x^2=0\\ \Rightarrow x\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)