Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

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

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

à xin lỗi mình nhầm dòng cuối 

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

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

Bài 1: 

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

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

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

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

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

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

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

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)

\(=a-2\sqrt{a}\)

help me

\(\left(x+3\right)^2+\left(x-2\right)^2-2\cdot\left(x+3\right)\left(x-2\right)\)\(=\left[\left(x+3\right)^2-2\cdot\left(x+3\right)\left(x-2\right)\right]+\left(x+2\right)^2\)\(=\left[x+3\cdot\left(x+3-2x+4\right)\right]+\left(x+2\right)^2\)

Tự làm tiếp.....

Ta có: \(\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)

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

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

\(=5^2=25\)

\(\left(x-2\right)^2-\left(x+2\right)\left(x-2\right)=\left(x^2-4x+4\right)-\left(x^2-4\right)=x^2-4x+4-x^2+4=8-4x\)

(x+2)²-(x+2)(x-2)

=x²+4x+4-x²+4

=4x+8

cho mình sửa đề xíu:   (x+2)^2-(x+2)(x-2)

\(=\dfrac{x+2}{\left(x-2\right)^2}:\left(\dfrac{6-x^2+x+x^2-4}{x\left(x-2\right)}\right)\)

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

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

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

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

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

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

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

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

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

giúp với aaaaaaaaaaaa

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

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

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

\(a=\dfrac{x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{1}{x^2}+\dfrac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

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

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