a) x= 2

b) x= 1

c) x= -1

d) x= 2

Mong mọi người duyệt câu hỏi của mình nha

2x-4=0

=>2x=4

=>x=2

x+x+2=O

=>2x=-2

=>x=-1

5-2x=x-1

=>-2x-x=-1-5

=>-3x=-6

=>x=3

3x-1=x-3

=>3x-x=-3+1

=>2x=-2

=>x=-1

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

 nửa chu vi của hình chữ nhật là: 60/2=30(cm)

chiều rộng cả hình chữ nhất là: 30-20=10(cm)

\(\left(x+20\right):2=60\)

\(\Rightarrow x+20=120\)

\(\Rightarrow x=100\)

a) A có nghĩa khi \(\hept{2x-2\ne02-2x^2\ne0\Leftrightarrow\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}\Leftrightarrow}x\ne\pm1}\)

Vậy A có nghĩa khi \(x\ne\pm1\)

b) \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\left(x\ne\pm1\right)\)

\(\Leftrightarrow A=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)

\(\Leftrightarrow\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow A=\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)

Vậy A=\(\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)

b) \(A=\frac{1}{2\left(x-1\right)}\left(x\ne\pm1\right)\)

A=\(\frac{-1}{2}\)\(\Leftrightarrow\frac{1}{2\left(x-1\right)}=\frac{-1}{2}\)

\(\Leftrightarrow-2\left(x-1\right)=2\)

<=> x-1=-1

<=> x=0 (tmđk)

Vậy x=0 thì \(A=\frac{-1}{2}\)

a) \(x\ne1,2;x\inℝ\)