Câu 2: 

a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)

\(=3x-\left|x-5\right|\)

\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)

b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

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

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

`@` `\text {Ans}`

`\downarrow`

\(B=(x+1)^2-2(2x-1)(1+x)+4x^2-4x+1\)

`= x^2 + 2x + 1 - 2(2x^2 + x - 1) + 4x^2 - 4x + 1`

`= 5x^2 - 2x + 2 - 4x^2 - 2x + 2`

`= x^2 - 4x + 4`

\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)

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

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

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

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

a: \(M=\dfrac{2x^2-10x-x^2+x+30-x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)

b: Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(x\in\left\{-4;-6;-3;-7;0;-10;-15\right\}\)

Sửa đề: \(B=\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}-\dfrac{5\sqrt{6}}{2}\)

Ta có: \(B=\dfrac{2}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}-\dfrac{5\sqrt{6}}{2}\)

\(=\dfrac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{5\sqrt{6}}{2}\)

\(=\dfrac{2\sqrt{6}+4+2\sqrt{6}-4}{2}-\dfrac{5\sqrt{6}}{2}\)

\(=-\dfrac{\sqrt{6}}{2}\)

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

b: Khi A=-3 thì \(\left|x\sqrt{2}+1\right|=x\sqrt{2}+3\)

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

\(\Leftrightarrow2x\sqrt{2}=-4\)

hay \(x=-\sqrt{2}\)