Câu hỏi của Mai Linh

Question

Cho biểu thức A = $\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)$

B = $\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}$ với x ≠ 9, x...

Vi Huyên · Accepted Answer

a) $A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)$

$A=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}$

$A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}$

$A=1$

b) Ta có:

$B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}$ ( x >= 0, x khác 9 )

$B=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{3+\sqrt{x}+3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{\left(3+\sqrt{x}\right)+3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{4\left(3+\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{4}{3-\sqrt{x}}$

Để B > A

$\Rightarrow\frac{4}{3-\sqrt{x}}>1$

$\Rightarrow4>3-\sqrt{x}$

$\Rightarrow4-3+\sqrt{x}>0$

$\Rightarrow1+\sqrt{x}>0$

$\Rightarrow\sqrt{x}>-1$

$\Rightarrow x>1$Câu trả lời: a) $A=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)$

$A=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}}$

$A=\frac{\sqrt{3}+1}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\frac{5+3\sqrt{5}}{\sqrt{5}}$

$A=1$

b) Ta có:

$B=\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}$ ( x >= 0, x khác 9 )

$B=\frac{3+\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{3+\sqrt{x}+3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{\left(3+\sqrt{x}\right)+3\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{4\left(3+\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$B=\frac{4}{3-\sqrt{x}}$

Để B > A

$\Rightarrow\frac{4}{3-\sqrt{x}}>1$

$\Rightarrow4>3-\sqrt{x}$

$\Rightarrow4-3+\sqrt{x}>0$

$\Rightarrow1+\sqrt{x}>0$

$\Rightarrow\sqrt{x}>-1$

$\Rightarrow x>1$

What ever · Answer

a) A=$\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)$

$=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}+\frac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}$

$=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}+\left(\sqrt{5}+3\right)-\left(\sqrt{5}+3\right)$

$=\frac{\sqrt{3}+1}{\sqrt{3}+1}+0=1$

b) B=$\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}$

$=\frac{3+\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{9-x}$

$=\frac{3+\sqrt{x}+3\sqrt{x}-x}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}$

$=\frac{4\text{​​}\sqrt{x}+12}{\left(3-\sqrt{x}\right)\cdot\left(3+\sqrt{x}\right)}$

$=\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}$

$=\frac{4}{3-\sqrt{x}}$

$B>A
\Leftrightarrow\frac{4}{3-\sqrt{x}}>1$

các giá trị của x là $\left\{x\in R\backslash0\le x\le9\right\}$Câu trả lời: a) A=$\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}+\frac{5+3\sqrt{5}}{\sqrt{5}}-\left(\sqrt{5}+3\right)$