a: ĐKXĐ: x<>1/2; x<>-1/2; x<>0

b: \(A=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{4x}\cdot\dfrac{5}{2x+1}=\dfrac{10}{2x+1}\)

tự làm đi

dễ vậy cũng hỏi

a)B =  \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)

= \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)

= \(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)

= \(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)

b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)

Thay x = -4 vào B, ta có:

B = \(\dfrac{-4.3}{-4+3}=12\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)

<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)

d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên

<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

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

b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)

Khi x=-1/2 thì B=2/5

c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

a, đk : x khác -2 ; 2 

\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

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

b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)

Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)

c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)

ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

Lỗi hình rồi em!

a.

\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\left(x\ge-1\right)\)

\(B=\sqrt{16}.\sqrt{x+1}-\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}+\sqrt{x+1}\)

\(B=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)

\(B=\left(4-3+2+1\right).\sqrt{x+1}\)

\(B=4.\sqrt{x+1}\)

b.

\(B=16\\\)

\(\Rightarrow4\sqrt{x+1}=16\)

\(\Rightarrow\sqrt{x+1}=\dfrac{16}{4}=4\)

\(\Rightarrow x+1=4^2\)

\(\Rightarrow x+1=16\rightarrow x=16-1=15\) (thỏa mãn)

vậy x=15

a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\left(\sqrt{x}+1\right)\)

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)