Question

[Phùng Ái Nguyên]

Ai giúp mình với ạ mình cần gấp (2 bài thôi cũng đc)

Answer 1

[Nguyễn Hoàng Minh]

Bài 8:

\(1,P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ 2,P=2\Leftrightarrow2\sqrt{x}+4=3\sqrt{x}\Leftrightarrow\sqrt{x}=4\\ \Leftrightarrow x=16\left(tm\right)\)

Bài 9:

\(a,M=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ M=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}-1\right)\\ M=\dfrac{x-1}{\sqrt{x}}\\ b,M>0\Leftrightarrow x-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\)

Bài 10:

\(a,A=\dfrac{\sqrt{\left(x+3\right)^2}}{x+3}=\dfrac{\left|x+3\right|}{x+3}\)

Với \(x\ge-3\Leftrightarrow A=\dfrac{x+3}{x+3}=1\)

Với \(x< -3\Leftrightarrow A=\dfrac{-\left(x+3\right)}{x+3}=-1\)

\(b,B=\dfrac{2}{x-1}\cdot\dfrac{\left|x-1\right|}{2\left|x\right|}\)

Với \(0< x< 1\Leftrightarrow B=\dfrac{2}{x-1}\cdot\dfrac{-\left(x-1\right)}{2x}=-\dfrac{1}{x}\)