a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

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

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)

c) \(A=\dfrac{1}{2}\) khi:

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

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

1/a)Ta có -5=-1*5=-5*1

​Sau đó bn lập bảng rồi dùng ước bội để tính nhé​

​ Tk mình nha bn!

Đáp án : D

1 

\(\left(x-2\right):2.3=6\)

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

\(\Leftrightarrow\left(x-2\right)=4\)

\(\Leftrightarrow x=4+2=6\)

c) ta có

\(\left[\left(2x+1\right)+1\right]m:2=625\)

\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)

\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)

\(\Leftrightarrow\left(2x+1\right)^2=1250\)

...

2

\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)

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

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

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

\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)

\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)

a) cho f(x) = 0

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

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

b)  2(x-3)-3(x+1)=5 

\(\Leftrightarrow2x-6-3x-3=5\)

\(\Leftrightarrow-x-9=5\)

\(\Leftrightarrow-x=14\Leftrightarrow x=14\)

`a)` Cho `f(x)=0`

`=>(x+2)(-x+1)=0`

`@TH1:x+2=0=>x=-2`

`@TH2:-x+1=0=>-x=-1=>x=1`

____________________________________________________

`b)2(x-3)-3(x+1)=5`

`=>2x-6-3x-3=5`

`=>2x-3x=5+6+3`

`=>-x=14`

`=>x=-14`

`a, 1/2 +x=3/4`

`=> x= 3/4 -1/2`

`=> x= 3/4-2/4`

`=>x= 1/4`

`b, 5/2 -x=1/3`

`=> x= 5/2 -1/3`

`=> x= 15/6 - 2/6`

`=>x= 13/6`

`c, 2 . (1/3 +x)=1/5`

`=> 1/3 +x=1/5:2`

`=> 1/3 +x= 1/10`

`=>x= 1/10-1/3`

`=>x= 3/30 - 10/30`

`=>x=-7/30`

`d, 2/3 - (1/2 -x)=1/5`

`=> 1/2-x= 2/3 -1/5`

`=>1/2-x= 10/15 - 3/15`

`=>1/2-x=7/15`

`=>x= 1/2-7/15`

`=>x=1/30`

`1/2 + x = 3/4`

`=>    x  = 3/4 - 1/2`

`=>    x   = 1/4`

`5/2 - x  = 1/3`

`=>    x  =  5/2 - 1/3`

`=>    x  = 13/6`

`2.(1/3 + x) = 1/5`

`=>1/3 + x  = 1/10 `

`=>         x =  1/10 - 1/3`

`=>        x   = -7/30`

`2/3 - (1/2 -x)= 1/5`

`=>     1/2 - x = 7/15`

`=>             x  = 1/2 - 7/15`

`=>             x  = 1/30`

a. \(\dfrac{1}{2}+x=\dfrac{3}{4}\)

⇔ \(x=\dfrac{3}{4}-\dfrac{1}{2}\)

⇔ \(x=\dfrac{1}{4}\)

b. \(\dfrac{5}{2}-x=\dfrac{1}{3}\)

⇔ \(-x=\dfrac{1}{3}-\dfrac{5}{2}\)

⇔ \(-x=-\dfrac{13}{6}\)

⇔ \(x=\dfrac{13}{6}\)

c. \(2\left(\dfrac{1}{3}+x\right)=\dfrac{1}{5}\)

⇔ \(\dfrac{1}{3}+x=\dfrac{1}{5}\div2\)

⇔ \(x=\dfrac{1}{10}-\dfrac{1}{3}\)

⇔ \(-\dfrac{7}{30}\)

d. \(\dfrac{2}{3}-\left(\dfrac{1}{2}-x\right)=\dfrac{1}{5}\)

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

⇔ \(x=-\dfrac{7}{15}+\dfrac{1}{2}\)

⇔ \(x=\dfrac{1}{30}\)

Đáp án D