# Ôn tập Căn bậc hai. Căn bậc ba

a) $\frac{1}{\sqrt{3}+\sqrt{2}}+\sqrt{7-4\sqrt{3}}+\sqrt{2}$ b) $\left(1+\frac{1}{cot^220^o}\right).cot^220^o-tan40^o.tan50^o$ c) A=$\frac{\sqrt{x}-1}{2\sqrt{x}+1}-\frac{3}{1-2\sqrt{x}}-\frac{4\sqrt{x}+4}{4x-1}$

Đõ Phương Thảo 29 tháng 11 2020 lúc 20:42

a) $\frac{1}{\sqrt{3}+\sqrt{2}}$ + $\sqrt{7-4\sqrt{3}}$ +$\sqrt{2}$

= $\frac{\sqrt{3}-\sqrt{2}}{3-2}$ + $\sqrt{4-2.2.\sqrt{3}+3}$+$\sqrt{2}$

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

= $\sqrt{3}-\sqrt{2}$+ $\left|2-\sqrt{3}\right|$+$\sqrt{2}$

= $\sqrt{3}-\sqrt{2}$ + 2-$\sqrt{3}$ + $\sqrt{2}$

=2.

b) $\left(1+\frac{1}{\cot^220^o}\right)$. $\cot^220^o$- $\tan40^o.\tan50^o$

=$\cot^220^o$ + 1 - $\tan40^o$ . $\cot40^o$

=$\cot^220^o$ + 1-1

= ​​$\cot^220^o$​.

c) A= $\frac{\sqrt{x}-1}{2\sqrt{x}+1}$ - $\frac{3}{1-2\sqrt{x}}$ - $\frac{4\sqrt{x}+4}{4x-1}$ , ĐK: x ≥ $\frac{1}{4}$

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

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

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

= $\frac{2x-\sqrt{x}-2\sqrt{x}+1+6\sqrt{x}+3-4\sqrt{x}-4}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}$

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

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

= $\frac{\sqrt{x}}{2\sqrt{x}+1}$.

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