a) 5-(x-6)=4.(3-2x)

<=>5-x+6=12-8x

<=>-x+8x=-5-6+12

<=>7x=1

<=>x=1/7

vậy nghiệm của phương trình là 1/7

b) 7-(2x+4)=-(x+4)

<=>7-2x-4=-x-4

<=>-2x+x=-7+4-4

<=>-x=-7

<=>x=7

vậy nghiệm của phương trình là 7

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

a) ĐKXĐ: x≠-5

Ta có: \(\dfrac{2x-5}{x+5}=4\)

\(\Leftrightarrow2x-5=4\left(x+5\right)\)

\(\Leftrightarrow2x-5=4x+20\)

\(\Leftrightarrow2x-5-4x-20=0\)

\(\Leftrightarrow-2x-25=0\)

\(\Leftrightarrow-2x=25\)

hay \(x=\dfrac{-25}{2}\)(nhận)

Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)

b) ĐKXĐ: x≠0

Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-8=2x^2+3x\)

\(\Leftrightarrow2x^2-8-2x^2-3x=0\)

\(\Leftrightarrow-3x-8=0\)

\(\Leftrightarrow-3x=8\)

hay \(x=\dfrac{-8}{3}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)

Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)

\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)

\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)

\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)

\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)

\(\Leftrightarrow20x+12=0\)

\(\Leftrightarrow20x=-12\)

hay \(x=-\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)

d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)

\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

a, 4x+1=13-2x <-->6x=12 <-->x=2

b, (2x-5)(x-4)=0 <-->x=5/2  hoặc x=4

c,Đề bài -->x(x-2)+6(x+2)=2x+12 -->x^2+2x=0 -->x=0  hoặc x=-2

d,|x-3|=9-2x -->TH1: x-3=9-2x -->x=x=4           TH2:3-x=9-2x -->x=6

\(a,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow2x+3=25\Leftrightarrow x=11\left(tm\right)\\ b,ĐK:x\ge2\\ PT\Leftrightarrow x^2+2x=2x+4\\ \Leftrightarrow x^2=4\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=2\)

\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

\(\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\)

\(ĐK:x\ne1;-3\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)+4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)

\(\Leftrightarrow13x=1\)

\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)

a)2.(x+3)-(3+x).(1`+2x)=0\(\Leftrightarrow\)2x+6-3-6x-x-2x\(^2\)=0

\(\Leftrightarrow\)-2x\(^2\)-5x+3=0\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-3;\dfrac{1}{2}\right\}\)

b)x\(^2\)-4x+4=9\(\Leftrightarrow\)x\(^2\)-4x+4-9=0\(\Leftrightarrow\)x\(^2\)-4x-5=0

\(\Leftrightarrow\left\{{}\begin{matrix}5-x=0\\1+x=0\end{matrix}\right.\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy PT đã cho có tập nghiệm S=\(\left\{-1;5\right\}\)

\(a,\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\-2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(b,\Leftrightarrow\left(x-2\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a) \(2\left(x+3\right)-\left(x+3\right)\left(1+2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2-1-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\1-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

-Vậy \(S=\left\{-3;\dfrac{1}{2}\right\}\)

b) \(x^2-4x+4=9\)

\(\Leftrightarrow\left(x-2\right)^2-9=0\)

\(\Leftrightarrow\left(x-2-3\right)\left(x-2+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
-Vậy \(S=\left\{5;-1\right\}\)

a, \(\left(x-5\right)\left(x-5+3\right)=0\Leftrightarrow x=5;x=2\)

b, \(-4x=\dfrac{274}{21}\Leftrightarrow x=-\dfrac{137}{42}\)

c, đk x khác - 2 ; 2 

\(x^2-3x+2-x^2-2x=6-7x\Leftrightarrow-5x+2=6-7x\)

\(\Leftrightarrow2x-4=0\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm