cái này mà tiếng anh à! Tào lao

Bài 1:

\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

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

Bài 2:

a) Không rõ đề

b) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)

1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.

\(\text{ĐKXĐ: }x-3\ge0;x+3\ge0;2x-6+\sqrt{x^2-9}\ne0\)

\(\Leftrightarrow x\ge3;x\ge-3;2x-6\ne\sqrt{x^2-9}\)

\(\Leftrightarrow x\ge3;4x^2-24x+36\ne x^2-9\)

\(\Leftrightarrow x\ge3;3x^2-24x+45\ne0\)

\(\Leftrightarrow x\ge3;3.\left(x^2-8x+15\right)\ne0\)

\(\Leftrightarrow x\ge3;\left(x-3\right)\left(x-5\right)\ne0\)

\(\Leftrightarrow x\ge3;x\ne3;x\ne5\)

\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}.\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-3\right)}}{2\left(x-3\right)+\sqrt{\left(x+3\right)\left(x-3\right)}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}.\sqrt{x+3}+2\sqrt{x+3}.\sqrt{x-3}}{2\sqrt{x-3}.\sqrt{x-3}+\sqrt{x+3}.\sqrt{x-3}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\sqrt{2}\)

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

\(\Leftrightarrow\frac{x+3}{x-3}=2\)

\(\Leftrightarrow x+3=2.\left(x-3\right)\)

\(\Leftrightarrow x+3=2x-6\)

\(\Leftrightarrow x-2x=-6-3\)

\(\Leftrightarrow-x=-9\)

\(\Leftrightarrow x=9\)

\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}.\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-3\right)}}{2\left(x-3\right)+\sqrt{\left(x+3\right)\left(x-3\right)}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}.\sqrt{x+3}+2\sqrt{x+3}.\sqrt{x-3}}{2\sqrt{x-3}.\sqrt{x-3}+\sqrt{x+3}.\sqrt{x-3}}=\sqrt{2}\)

\(\Leftrightarrow\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\sqrt{2}\)

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

\(\Leftrightarrow\frac{x+3}{x-3}=2\)

\(\Leftrightarrow x+3=2.\left(x-3\right)\)

\(\Leftrightarrow x+3=2x-6\)

\(\Leftrightarrow x-2x=-6-3\)

\(\Leftrightarrow-x=-9\)

\(\Leftrightarrow x=9\)

1/ Đặt \(\sqrt{9-x^2}=a\ge0\)

\(\Rightarrow\frac{9-a^2}{3+a}+\frac{1}{12-4a}=1\)

\(\Leftrightarrow4a^2-20a+25=0\)

\(\Leftrightarrow a=\frac{5}{2}\)

\(\Rightarrow\sqrt{9-x^2}=\frac{5}{2}\)

\(\Leftrightarrow x^2=\frac{11}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{11}}{2}\\x=\frac{\sqrt{11}}{2}\end{cases}}\)

2/ \(\frac{9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-1=0\)

\(\Leftrightarrow\frac{9+2x^2}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-3=0\)

Đặt \(\frac{x}{\sqrt{2x^2+9}}=a\)

\(\Rightarrow\frac{1}{a^2}+2a-3=0\)

\(\Leftrightarrow2a^3-3a^2+1=0\)

\(\Leftrightarrow\left(a-1\right)^2\left(2a+1\right)=0\)

Làm nốt nhé

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

\(\Leftrightarrow x-\sqrt{2}+\frac{x-\sqrt{2x^2-2}}{\sqrt{x^2-1}}=0\)

\(\Leftrightarrow x-\sqrt{2}+\frac{2-x^2}{\sqrt{x^2-1}.\left(x+\sqrt{2x^2-2}\right)}=0\)

\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(1+\frac{\sqrt{2}+x}{\sqrt{x^2-1}.\left(x+\sqrt{2x^2-2}\right)}\right)=0\)

\(\Leftrightarrow x=\sqrt{2}\)