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a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
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/ \(7x-5=13-5x\)
\(\Leftrightarrow7x+5x=13+5\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\frac{3}{2}\)
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!
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
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