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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 \(S=\left\{100\right\}\)
\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{x+1}{x-2}-\frac{1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2+3x+3}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow x^2+3x+3-1=0\)
\(\Leftrightarrow x^2+3x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
<=> x+1=0 hoặc x+2=0
<=> x=-1 hoặc x=-2
\(b,\frac{3}{x+1}=\frac{5}{2x+2}\)
\(\frac{3}{x+1}=\frac{5}{2\left(x+1\right)}\)
\(3=\frac{5}{2}\left(vl\right)\)vô nghiệm
Giải phương trình:
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
\(\Leftrightarrow\left(\frac{x+1}{58}+1\right)+\left(\frac{x+2}{57}+1\right)=\left(\frac{x+3}{56}+1\right)+\left(\frac{x+4}{55}+1\right)\)
\(\Leftrightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)
\(\Leftrightarrow x+59=0\) \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\right)\)
\(\Leftrightarrow x=-59\)
Vậy : \(S=\left\{-59\right\}\)
\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)
\(\Leftrightarrow\) \(\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)
\(\Leftrightarrow\) \(\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)
\(\Leftrightarrow\) \(\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)
\(\Leftrightarrow\) (x + 59)(\(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\)) = 0
\(\Leftrightarrow\) x + 59 = 0
\(\Leftrightarrow\) x = -59
Vậy S = {-59}
Chúc bn học tốt!!
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)
\(\frac{1}{x}-\frac{x+2}{x-2}=\frac{2}{x\cdot\left(2-x\right)}\\ \Leftrightarrow\frac{1}{x}-\frac{x-2}{2-x}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x}{x\cdot\left(2-x\right)}-\frac{x^2-2x}{x\cdot\left(2-x\right)}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x-x^2+2x-2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{x-x^2}{x\cdot\left(2-x\right)}=0\\ \Rightarrow x-x^2=0\\ \Rightarrow x\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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 : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)
Phương trình nhận \(x=2\)làm nghiệm nên :
\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)
\(\Leftrightarrow15m+90-20=80\)
\(\Leftrightarrow15m=80+20-90\)
\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)
....
b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)
Phương trình nhận \(x=1\)làm nghiệm nên :
\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)
\(\Leftrightarrow30+15m-32=43\)
\(\Leftrightarrow15m=43+32-30\)
\(\Leftrightarrow15m=45\Leftrightarrow m=3\)
....
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80
Phương trình có nghiệm x = 2:
5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80
<=> 5(m + 6).3 - 4.5 = 80
<=> 15(m + 6) - 4.5 = 80
<=> 15(m + 6) - 20 = 80
<=> 15(m + 6) = 80 + 20
<=> 15(m + 6) = 100
<=> m + 6 = 100 : 15
<=> m + 6 = 20/3
<=> m = 20/3 - 6
<=> m = 2/3
b) 3(2x + m)(3x + 2) - 2(3x + 1)2 = 43
Phương trình có nghiệm x = 1:
3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)2 = 43
<=> 3(2 + m).5 - 2.16 = 43
<=> 15(2 + m) - 32 = 43
<=> 15(2 + m) = 43 + 32
<=> 15(2 + m) = 75
<=> 2 + m = 75 : 15
<=> 2 + m = 5
<=> m = 5 - 2
<=> m = 3
Bạn xem lại xem có viết nhầm đề bài không thế?