Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

ghê thậc, còn cái còn lại thì seo?

a.

ĐKXĐ: \(x\ge0\)

\(\sqrt{2x^2+13x+5}-5\sqrt{x}+\sqrt{2x^2-3x+5}-3\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2-12x+5}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{2x^2-12x+5}{\sqrt{2x^2-3x+5}+3\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-12x+5\right)\left(\dfrac{1}{\sqrt{2x^2+13x+5}+5\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-3x+5}+3\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-12x+5=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x^2\ge\dfrac{4}{3}\)

\(\sqrt{x^2-\dfrac{4}{3}}+\sqrt{4x^2-4}-x=0\)

\(\Leftrightarrow\sqrt{\dfrac{3x^2-4}{3}}+\dfrac{3x^2-4}{\sqrt{4x^2-4}+x}=0\)

\(\Leftrightarrow\sqrt{3x^2-4}\left(\dfrac{1}{\sqrt{3}}+\dfrac{\sqrt{3x^2-4}}{\sqrt{4x^2-4}+x}\right)=0\)

\(\Leftrightarrow3x^2-4=0\)

\(\Leftrightarrow...\)

Thắng Chó Râm tặc

\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)

\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

ĐKXĐ: 5-x>=0 và x+8>=0

=>-8<=x<=5

Ta có: \(13\sqrt{5-x}+18\sqrt{x+8}=61+x+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\sqrt{5-x}-26+18\sqrt{x+8}-54=x-19+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\left(\sqrt{5-x}-2\right)+18\left(\sqrt{x+8}-3\right)=x-1-18+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\frac{5-x-4}{\sqrt{5-x}+2}+18\cdot\frac{x+8-9}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{\left(5-x\right)\left(x+8\right)}-6\right)\)

=>\(13\cdot\frac{1-x}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{5x+40-x^2-8x}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{-x^2-3x+40}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\cdot\frac{-x^2-3x+40-36}{\sqrt{-x^2-3x+40}+6}\)

=>(x-1)\(\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-3x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-4x+x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{\left(x+4\right)\left(-x+1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1-3\cdot\frac{\left(x+4\right)\left(x-1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}-1+3\cdot\frac{x+4}{\sqrt{-x^2-3x+40}+6}\right)=0\)

=>x-1=0

=>x=1(nhận)

\(\sqrt{x^{ }2-6x+9}=4-x\)
\(\sqrt{\left(x-3\right)^{ }2}=4-x\)
x-3=4-x
x+x=4+3
2x=7
x=\(\dfrac{7}{2}\)

Lời giải:
a.

PT \(\Leftrightarrow \left\{\begin{matrix} 4-x\geq 0\\ x^2-6x+9=(4-x)^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 4\\ 2x=7\end{matrix}\right.\Leftrightarrow x=\frac{7}{2}\)

b.

ĐKXĐ: $x\geq \frac{3}{2}$

PT \(\Leftrightarrow \sqrt{(2x-3)+2\sqrt{2x-3}+1}+\sqrt{(2x-3)+8\sqrt{2x-3}+16}=5\)

\(\Leftrightarrow \sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5\)

\(\Leftrightarrow |\sqrt{2x-3}+1|+|\sqrt{2x-3}+4|=5\)

\(\Leftrightarrow \sqrt{2x-3}+1+\sqrt{2x-3}+4=2\sqrt{2x-3}+5=5\)

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