ĐKXĐ: \(0\le x\le5\).

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\).

PT đã cho tương đương với: \(\left(8-ab\right)\left(a-b\right)=2\left(a-b\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=6\end{matrix}\right.\).

+) \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=2,5\left(TMĐK\right)\).

+) \(ab=6\Leftrightarrow\sqrt{x\left(5-x\right)}=6\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\).

Vậy...

ĐK: \(0\le x\le5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(8-ab\right)\left(a-b\right)=2\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(a-b\right)\left(8-ab-2a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab+2a+2b=8\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH2: \(ab+2a+2b=8\)

\(\Leftrightarrow\sqrt{5x-x^2}+2\sqrt{5-x}+2\sqrt{x}=8\)

\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x}-3\right)\left(\sqrt{5-x}+\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5-x}+\sqrt{x}=-7\left(l\right)\\\sqrt{5-x}+\sqrt{x}=3\end{matrix}\right.\)

\(\sqrt{5-x}+\sqrt{x}=3\)

\(\Leftrightarrow5+2\sqrt{5x-x^2}=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Vậy ...

mình nghĩ sửa đề bài là  \(\frac{\sqrt{x^2-x+6}+7\sqrt{x}-\sqrt{6\left(x^2+5x-2\right)}}{x+3-\sqrt{2\left(x^2+10\right)}}\le0\) 

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=4-\sqrt{5}+\sqrt{5}+1=5\)

Bài 2:

a: ĐKXĐ: x>=3

\(\sqrt{x-3}=6\)

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)

\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)

\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)

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

b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

\(=\sqrt{3x-1}+\sqrt{5}\)

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{2\left(a-2\right)}{a+2}\)

Đặt:\(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(N\right)\end{matrix}\right.ĐKXĐ:x>=-6\)

\(S=\left\{16,-5\right\}\)

Câu trên mình quên -5>-6

Đặt: \(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\)

\(\Leftrightarrow v+5u-5-uv=0\)

\(\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\)

\(\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(L\right)\end{matrix}\right.\)          ĐKXĐ:\(x>=-6\)

\(S=\left\{16\right\}\)

\(\sqrt{x+9}+5\sqrt{x+\text{6}}=5+\sqrt{\left(x+9\right)\left(x+\text{6}\right)}\Leftrightarrow\sqrt{x+9}+5=5+\sqrt{x+9}\Leftrightarrow\sqrt{x+9}-\sqrt{x+9}=0\Leftrightarrow x+9-x-9=0\Leftrightarrow0=0\)

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