Question

[Hoàng Khánh Linh]

Mọi người giúp em với ạ

Answer 1

[Nguyễn Huy Tú]

Bài 3 : 

a, Với \(x\ge0;x\ne1\)

\(P=\frac{3x-\sqrt{x}-8}{x+\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+2}\)

\(=\frac{3x-\sqrt{x}-8-\left(x-4\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x-\sqrt{x}-8-x+4+2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2\sqrt{x}-3}{\sqrt{x}-1}\)

b, Ta có : \(x=\frac{4}{\sqrt{7}+\sqrt{5}}+\frac{6}{2-\sqrt{7}}+10=\frac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}+\frac{6\left(2+\sqrt{7}\right)}{-3}+10\)

\(=2\sqrt{7}-2\sqrt{5}-4-2\sqrt{7}+10=-2\sqrt{5}+6\)

\(\Rightarrow\sqrt{x}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

Thay vào P ta được : \(\frac{2\left(\sqrt{5}-1\right)-3}{\sqrt{5}-1-1}=\frac{2\sqrt{5}-5}{\sqrt{5}-2}=-\sqrt{5}\)

Answer 2

[Nguyễn Huy Tú]

c, Ta có : \(\frac{2\sqrt{x}-3}{\sqrt{x}-1}\le1\Leftrightarrow\frac{2\sqrt{x}-3}{\sqrt{x}-1}-1\le0\)

\(\Leftrightarrow\frac{2\sqrt{x}-3-\sqrt{x}+1}{\sqrt{x}-1}\le0\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}-1}\le0\)

Vì \(\sqrt{x}-1>\sqrt{x}-2\)

\(\hept{\begin{cases}\sqrt{x}-2\le0\\\sqrt{x}-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le4\\x\ge1\end{cases}}\Leftrightarrow1\le x\le4\)

Kết hợp với đk vậy \(1< x\le4\)

d, Ta có : \(\frac{2\sqrt{x}-3}{\sqrt{x}-1}-\frac{3}{2}\le0\Leftrightarrow\frac{4\sqrt{x}-6-3\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}\le0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}-1\right)}\le0\) TH1 : \(\hept{\begin{cases}\sqrt{x}-3\le0\\\sqrt{x}-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le9\\x\ge1\end{cases}}\Leftrightarrow1< x\le9\)

TH1 : \(\hept{\begin{cases}\sqrt{x}-3\ge0\\\sqrt{x}-1\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge9\\0\le x< 1\end{cases}}\)( vô lí )