tui cx có bài như vạy nhưng 50 là 17 cơ, đặt 2 cái đầu =a, sau đó tìm a^2

Ta có \(A^2=\left(x-\sqrt{50}+x-\sqrt{50}-2.\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=\left(2x-2.\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=2.\left(x-\sqrt{x^2-50}\right).\left(x+\sqrt{x^2-50}\right)\)

\(=2.\left(x^2-x^2+50\right)\)

\(=100\)

Ta có \(\sqrt{x-\sqrt{50}}< \sqrt{x+\sqrt{50}}\)

\(\Rightarrow\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}< 0\)

mà \(\sqrt{x+\sqrt{x^2-50}}\ge0\)

Nên \(A\le0\)

Có \(A^2=100\)

Nên A=-10

\(\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\sqrt{x+\sqrt{x^2-50}}\)

\(=\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right).\frac{1}{\sqrt{2}}.\sqrt{2x+2\sqrt{x-\sqrt{50}}.\sqrt{x+\sqrt{50}}}\)

\(=\frac{1}{\sqrt{2}}.\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\sqrt{\left(\sqrt{x-\sqrt{50}}+\sqrt{x+\sqrt{50}}\right)^2}\)

\(=\frac{1}{\sqrt{2}}.\left(\sqrt{x-\sqrt{50}}-\sqrt{x+\sqrt{50}}\right)\left(\sqrt{x-\sqrt{50}}+\sqrt{x+\sqrt{50}}\right)\)

\(=\frac{1}{\sqrt{2}}.\left(x-\sqrt{50}-x-\sqrt{50}\right)=\frac{-2\sqrt{50}}{\sqrt{2}}=-10\)

a. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

$\Leftrightarrow \sqrt{2x}=3$

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

PT $\Leftrightarrow |2x-1|-x=3$

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

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

f.

ĐKXĐ: $x\geq 2$

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

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

$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$

$\Leftrightarrow x=2$ hoặc $x=5$ (tm)

h. ĐKXĐ: $x\leq \frac{3}{2}$

PT $\Leftrightarrow \sqrt{3-2x}=x+2$

\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)

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

Vậy.......

a: =>\(x\cdot\left(\sqrt{3}-1\right)=16\)

=>\(x=\dfrac{16}{\sqrt{3}-1}=8\left(\sqrt{3}+1\right)\)

b: =>(x-căn 15)^2=0

=>x-căn 15=0

=>x=căn 15

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)

cái này là Toán lớp 2?

\(3x-2=\sqrt[]{x^2+15}-\sqrt[]{x^2+8}=\dfrac{7}{\sqrt[]{x^2+15}+\sqrt[]{x^2+8}}>0\)

\(\Rightarrow x>\dfrac{2}{3}\)

\(\sqrt[]{x^2+15}-4=3x-3+\sqrt[]{x^2+8}-3\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+15}+4}=3\left(x-1\right)+\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+8}+3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{x+1}{\sqrt[]{x^2+15}+4}=3+\dfrac{x+1}{\sqrt[]{x^2+8}+3}\left(1\right)\end{matrix}\right.\)

Do \(x>\dfrac{2}{3}\Rightarrow x+1>0\Rightarrow\dfrac{x+1}{\sqrt[]{x^2+15}+4}< \dfrac{x+1}{\sqrt[]{x^2+8}+3}\)

\(\Rightarrow\) (1) vô nghiệm hay pt có nghiệm duy nhất \(x=1\)

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)