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\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
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= \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
=\(\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)
= \(\frac{\sqrt{3}+3+\sqrt{2}-\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{2}+\sqrt{5}\right)}\)
= \(\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{5}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}}\)
= \(\sqrt{3}+\sqrt{5}+3\)
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\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{5}-\sqrt{7}-1\)
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a) S=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}.\)
\(\sqrt{2}.\)S=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}.\)
\(\sqrt{2}.\)S =\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(\sqrt{2}.\)S=|\(\sqrt{7}\)-1|+|\(\sqrt{7}\)+1|=\(\sqrt{7}\)-1-\(\sqrt{7}\)-1=- 2
S= - \(\sqrt{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
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a) \(M=\sqrt[3]{7+5\sqrt{2}}\)
Ta có:
Vì \(7+5\sqrt{2}=\left(\sqrt{2}\right)^3+1+3\sqrt{2}.1\left(\sqrt{2}+1\right)=\left(\sqrt{2}+1\right)^3\)
Nên \(M=\sqrt[3]{\left(\sqrt{2}+1\right)^3}=\sqrt{2}+1\)
b) \(N=\sqrt[3]{6\sqrt{3}-10}\)
Ta có:
Vì \(6\sqrt{3}-10=\left(\sqrt{3}\right)^3-1^3-3\sqrt{3}.1\left(\sqrt{3}-1\right)=\left(\sqrt{3}-1\right)^3\)
Nên \(N=\sqrt[3]{\left(\sqrt{3}-1\right)^3=\sqrt{3}-1}\)
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\(B=\frac{\sqrt{3}+\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}}{\sqrt{2}+\sqrt{5+2\sqrt{5}.1+1}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\)
\(=\frac{3}{1}\)\(=3\)
\(=\frac{\sqrt{3}+\text{ |3+\sqrt{2}|}-\text{ |\sqrt{3}+\sqrt{2}|}}{\sqrt{2}+\text{ |\sqrt{5}+1| - \text{ |\sqrt{5}+\sqrt{2}|}}}\)
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Đặt tử = A , MẪu = B
=> A = \(\sqrt{3}+\sqrt{9+2.3\sqrt{2}+2}-\sqrt{2+2\sqrt{2}.\sqrt{3}+3}=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}=3\)
Tuwg tự tính B
ta có B = 1
A/B = 3/1 = 3
Nhớ ấn đúng nha
\(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5-2.\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}+1=-\sqrt{2}+1\)