\(x-4\sqrt{x}-6=0\)

\(< =>\sqrt{x}^2-4\sqrt{x}-6=0\)

\(\left(a=1;b=-4;b'=-2;c=-6\right)\)

\(\Delta'=b'^2-ac\)

    \(=\left(-2\right)^2-1.\left(-6\right)\)

   \(=4+6\)

   \(=10>0\)

\(\sqrt{\Delta'}=\sqrt{10}\)

Phương trình có 2 nghiệm phân biệt 

\(\sqrt{x_1}=\frac{2+\sqrt{10}}{1}=2+\sqrt{10}\)

\(\sqrt{x_2}=\frac{2-\sqrt{10}}{1}=2-\sqrt{10}\)

Với \(\sqrt{x_1}=2+\sqrt{10}\) suy ra \(x_1=\left(2+\sqrt{10}\right)^2=14+4\sqrt{10}\)

Với \(\sqrt{x_2}=2-\sqrt{10}\) suy ra \(x_2=\left(2-\sqrt{10}\right)^2=14-4\sqrt{10}\)

HỌC TỐT !!! 

a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)

đk: x >/ 0

(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)

Kl: \(x=\dfrac{392}{169}\)

b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)

đk: x >/ 5

(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)

Kl: x=9

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

Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)

(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)

Kl: x=-6

d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)

Đk: \(x\ge\dfrac{4}{5}\)

(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)

Kl: x=12

a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}

\(\sqrt{3x-5}=\sqrt{7x-1}\)

\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)

\(\left|3x-5\right|=\left|7x-1\right|\)

\(3x-5=7x-1\)

\(-4x=4\) => x = -1

đang vội nên mk làm tắt nha . đk x>=-5/4

\(\Leftrightarrow2\left(x+1\right)\)\(.\left[\left(x+2\right)-\sqrt{4x+5}\right]+2 \left(x+5\right)\sqrt{x+3}\left(\sqrt{x+3}-2\right)+\)\(2x^2+6x-8=0\)

\(\Leftrightarrow\frac{2\left(x+1\right)^2\left(x-1\right)}{x+2+\sqrt{4x+5}}+\frac{2\left(x+5\right)\left(x-1\right)\sqrt{x+3}}{\sqrt{x+3}+2}+2\left(x-1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{2\left(x+1\right)^2}{x+2+\sqrt{4x+5}}+\frac{2\left(x+5\right)\sqrt{x+3}}{\sqrt{x+3}+2}+2\left(x+4\right)\right]=0\)

de thấy bt trong ngoặc dương suy ra x=1 là no

\(a,x-3\sqrt{x}+2\)

\(=x-3\sqrt{x}+\frac{9}{4}-\frac{1}{4}\)

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

câu a mình nhìn nhầm :

\(=\left(x-1\right)\left(x+2\right)\)

a) \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow-7\sqrt{x}=0-3x-4\)

Bình phương hai vế, ta có:

\(\Leftrightarrow49x=9x^2+24x+16\)

\(\Leftrightarrow49x-9x^2-24x-16=0\)

\(\Leftrightarrow25x-9x^2-16=0\)

\(\Leftrightarrow9x^2-25x+16=0\)

\(\Leftrightarrow9x^2-9x-16x+16=0\)

\(\Leftrightarrow9x\left(x-1\right)-16\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(9x-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\9x-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}\)

vậy nghiệm phương trình là: \(\left\{1;\frac{16}{9}\right\}\)

b) bình phương 2 vế và làm tương tự, mình hơi lười