Question

Cho \(B=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)

a) Tìm ĐKXĐ và rút gọn

b)CM \(B=\frac{2}{a-1}\)

Đừng làm tắt :<

Answer 1

a, \(B=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)ĐKXĐ : \(a>0;a\ne1\)

\(=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right)\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{a-1}\right)\frac{1}{\sqrt{a}}\)

\(=\frac{2\sqrt{a}}{\left(a-1\right)\sqrt{a}}=\frac{2}{a-1}\)

b, quá rõ ràng rồi nhé 

Answer 2

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x(x+1)(x&#x2212;x+1)x&#x2212;x+1&#x2212;x(2x+1)x+1" role="presentation" style="border:0px; box-sizing:inherit; direction:ltr; display:inline-block; float:none; font-size:18.08px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">x(x+1)(x−x+1)x−x+1−x(2x+1)x+1

$\iff A=x+\sqrt{x}-2\sqrt{x}-1+1$

$\iff A=x-\sqrt{x}$

b) Để A = 0

$\iff x-\sqrt{x}=0$

$\iff \sqrt{x}\left(\sqrt{x}-1\right)=0$

⇔[√x=0√x=1⇔[x=0x=1

⇔[x=0(ktm)x=1(tm)⇔[x=0(ktm)x=1(tm)

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