Có mỗi cái yêu cầu ở đầu bài, còn phần cần thiết nhất thì lại không có. Tóm lại bạn có câu hỏi gì?

\(\sqr{3} + {2} \sqr{2} - \sqr{3} - 2 \sqr{2}\)

\(\sqr{7 } -{4}\sqr3 + \sqr{4} + 4 \sqr{4}\)

\(\sqr{23}+ {8} \sqr{7} - \sqr{7}\)

\(\sqr{11} - 6\sqr{2} + {3} + \sqr{2}\)

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

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

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

câu a mình nhìn nhầm :

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

b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)

c: \(=\left|x-4\right|+\left|x-6\right|\)

=x-4+6-x=2

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

Bài 1:

a) Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

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

\(=x-1\)

b) Ta có: \(\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

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

\(=x\sqrt{x}+1\)

c) Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)

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

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

Bài 2: Tìm x

a) Ta có: \(\sqrt{9x^2+6x+1}=3x-2\)

\(\Leftrightarrow\left|3x+1\right|=3x-2\)(*)

Trường hợp 1: \(x\ge\frac{-1}{3}\)

(*)\(\Leftrightarrow3x+1=3x-2\)

\(\Leftrightarrow3x+1-3x+2=0\)

\(\Leftrightarrow3=0\)(vô lý)

Trường hợp 2: \(x< \frac{-1}{3}\)

(*)\(\Leftrightarrow-3x-1=3x-2\)

\(\Leftrightarrow-3x-1-3x+2=0\)

\(\Leftrightarrow-6x+1=0\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\frac{1}{6}\)(loại)

Vậy: \(S=\varnothing\)

b)Trường hợp 1: \(x\ge0\)

Ta có: \(\sqrt{x}-2>0\)

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

hay x>4(nhận)

Vậy: S={x|x>4}

Cảm ơn ạ

@Hắc Hường