Question

[Bùi Xuân Chiến]

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giải hộ vs ạ

Answer 1

[Nguyễn Minh Quang]

các bài khá giống nhau nên mình làm 1 bài thôi nhé

bài 2.

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

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

ta có : \(A=1+\frac{2}{\sqrt{x}-1}\text{nguyên}\Rightarrow\sqrt{x}-1\in\left\{-1,1,2\right\}\Rightarrow x\in\left\{0,4,9\right\}\)

kết hợp điều kiện ta có \(x\in\left\{4,9\right\}\)

Answer 2

[Nguyễn Huy Tú]

Bài 1 : 

a, Với \(x\ge0;x\ne9;25\)

\(P=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\frac{25-x-\left(x-9\right)+\left(x-25\right)}{x+2\sqrt{x}-25}\right)\)

\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{x+2\sqrt{x}-25}\right)=-\frac{5}{\sqrt{x}+5}:\left(\frac{-x+9}{x+2\sqrt{x}-25}\right)\)

\(=-\frac{5}{\sqrt{x}+5}:\frac{-\left(x-9\right)}{x+2\sqrt{x}-25}=-\frac{5}{\sqrt{x}+5}:\frac{-\sqrt{x}-3}{\sqrt{x}+5}=\frac{5}{\sqrt{x}+3}\)

b, \(P< 1\Rightarrow\frac{5}{\sqrt{x}+3}-1< 0\Leftrightarrow\frac{5-\sqrt{x}-3}{\sqrt{x}+3}< 0\)

\(\Rightarrow2-\sqrt{x}< 0\Leftrightarrow-\sqrt{x}< -2\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

Kết hợp với đk vậy \(x>4;x\ne9;25\)