nhanh mik tích cho

help me

\(a,\)

\(=4x^2+4x-2xy-2y+x^2+2x+1+4x^2-4xy-y^2\)

\(=9x^2+6x-6xy-2y-y^2+1\)

\(b,\)

\(\dfrac{2x^2+3x-2}{2x-1}=\dfrac{\left(2x^2-x\right)+\left(4x-2\right)}{2x-1}\)

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

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

Thay x = 2 ; y = -1 ta được 

\(A=4-3\left(-1\right)+4.4=4+3+16=23\)

ĐKXĐ:\(x\ne-1,x\ne-2\)

\(\dfrac{\left(x+2\right)\left(2x-1\right)-x-2}{x^2+3x+2}=0\\ \Rightarrow2x^2+4x-x-2-x-2=0\\ \Leftrightarrow2x^2+2x-4=0\\ \Leftrightarrow x^2+x-2=0\\ \Leftrightarrow\left(x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Vậy \(x\in\left\{1\right\}\)

Chọn D

C

C nha 

Lời giải:
\(B=\frac{2x+\sqrt{x}-4}{(\sqrt{x}+1)(\sqrt{x}-2)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+1)(\sqrt{x}-2)}+\frac{\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-2)}\)

\(=\frac{2x+\sqrt{x}-4-(x-4)+\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-2)}=\frac{x+2\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-2)}=\frac{(\sqrt{x}+1)^2}{(\sqrt{x}+1)(\sqrt{x}-2)}=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(P=AB=\frac{3\sqrt{x}-4}{\sqrt{x}+1}.\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{3\sqrt{x}-4}{\sqrt{x}-2}\)

\(P\geq 2\Leftrightarrow \frac{3\sqrt{x}-4}{\sqrt{x}-2}\geq 2\)

\(\Leftrightarrow \frac{3\sqrt{x}-4}{\sqrt{x}-2}-2\geq 0\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}-2}\geq 0\)

\(\Leftrightarrow \sqrt{x}-2>0\Leftrightarrow x>4\)

Kết hợp với ĐKXĐ suy ra $x>4$

lên google mà xem đi chòi

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huhuhuhuh

tk: