Câu hỏi của Nguyễn Bảo Ánh

Question

Mọi người giúp em ạ undefined

Lê Song Phương · Accepted Answer

a) đkxđ: $\hept{\begin{cases}x\ge0\x\ne1\end{cases}}$

$P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}+x+2}{1-x}\right):$$\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{x-3+\sqrt{x}}{x-\sqrt{x}}\right)$$=\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+x+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]$$:\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x-3+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]$

$=\left[\frac{x-\sqrt{x}-\sqrt{x}-x-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]$$:\left[\frac{x-1-x+3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]$

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

$=\frac{-2\left(\sqrt{x}+1\right).\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(2-\sqrt{x}\right)}$$=\frac{-2\sqrt{x}}{2-\sqrt{x}}$
b) Để P = -1 thì $\frac{-2\sqrt{x}}{2-\sqrt{x}}=-1\Leftrightarrow\frac{2\sqrt{x}}{2-\sqrt{x}}=1\Rightarrow2\sqrt{x}=2-\sqrt{x}\Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}$(nhận)

c) Để P < -1/2 thì $\frac{-2\sqrt{x}}{2-\sqrt{x}}< \frac{-1}{2}\Leftrightarrow\frac{2\sqrt{x}}{2-\sqrt{x}}>\frac{1}{2}\Rightarrow4\sqrt{x}>2-\sqrt{x}\Leftrightarrow5\sqrt{x}>2\Leftrightarrow\sqrt{x}>\frac{2}{5}\Leftrightarrow x>\frac{4}{25}$

Vậy để P < -1/2 thì $\hept{\begin{cases}x>\frac{4}{25}\x\ne1\end{cases}}$

d) Ta có $x=4+2\sqrt{3}=3+2\sqrt{3}+1=\left(\sqrt{3}\right)^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2$

Thay $x=\left(\sqrt{3}+1\right)^2$vào P, ta có:

$P=\frac{-2\sqrt{x}}{2-\sqrt{x}}=\frac{-2\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{-2\left(\sqrt{3}+1\right)}{2-\sqrt{3}-1}=\frac{-2\left(\sqrt{3}+1\right)}{1-\sqrt{3}}$$=\frac{-2\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}=\frac{-2\left(4+2\sqrt{3}\right)}{1-3}=4+2\sqrt{3}$

Câu trả lời:

a) đkxđ: $\hept{\begin{cases}x\ge0\x\ne1\end{cases}}$

$P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}+x+2}{1-x}\right):$$\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{x-3+\sqrt{x}}{x-\sqrt{x}}\right)$$=\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+x+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]$$:\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x-3+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]$

$=\left[\frac{x-\sqrt{x}-\sqrt{x}-x-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]$$:\left[\frac{x-1-x+3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]$

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

$=\frac{-2\left(\sqrt{x}+1\right).\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(2-\sqrt{x}\right)}$$=\frac{-2\sqrt{x}}{2-\sqrt{x}}$
b) Để P = -1 thì $\frac{-2\sqrt{x}}{2-\sqrt{x}}=-1\Leftrightarrow\frac{2\sqrt{x}}{2-\sqrt{x}}=1\Rightarrow2\sqrt{x}=2-\sqrt{x}\Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}$(nhận)

c) Để P < -1/2 thì $\frac{-2\sqrt{x}}{2-\sqrt{x}}< \frac{-1}{2}\Leftrightarrow\frac{2\sqrt{x}}{2-\sqrt{x}}>\frac{1}{2}\Rightarrow4\sqrt{x}>2-\sqrt{x}\Leftrightarrow5\sqrt{x}>2\Leftrightarrow\sqrt{x}>\frac{2}{5}\Leftrightarrow x>\frac{4}{25}$

Vậy để P < -1/2 thì $\hept{\begin{cases}x>\frac{4}{25}\x\ne1\end{cases}}$

d) Ta có $x=4+2\sqrt{3}=3+2\sqrt{3}+1=\left(\sqrt{3}\right)^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2$

Thay $x=\left(\sqrt{3}+1\right)^2$vào P, ta có:

$P=\frac{-2\sqrt{x}}{2-\sqrt{x}}=\frac{-2\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{-2\left(\sqrt{3}+1\right)}{2-\sqrt{3}-1}=\frac{-2\left(\sqrt{3}+1\right)}{1-\sqrt{3}}$$=\frac{-2\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}=\frac{-2\left(4+2\sqrt{3}\right)}{1-3}=4+2\sqrt{3}$

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