So sánh \(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\) và \(B=\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{20}+\sqrt{40}+\sqrt{60}\right)\)
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Lời giải:
a)
\(\frac{2A}{\sqrt{2}}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{3+1+2\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3}}}+\frac{3+1-2\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3}}}\)
\(=\frac{(\sqrt{3}+1)^2}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{(\sqrt{3}-1)^2}{2-\sqrt{(\sqrt{3}-1)^2}}=\frac{(\sqrt{3}+1)^2}{2+\sqrt{3}+1}+\frac{(\sqrt{3}-1)^2}{2-(\sqrt{3}-1)}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{3}(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)^2}{\sqrt{3}(\sqrt{3}-1)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}=2\)
$\Rightarrow A=\sqrt{2}$
b)
\(B=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}=\sqrt{(8+2\sqrt{15})+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
c)
\(C=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}=\frac{\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+2\sqrt{x-1}+1}}{\sqrt{(x-2)^2}}\)
\(=\frac{\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}}{|x-2|}=\frac{|\sqrt{x-1}-1|+|\sqrt{x-1}+1|}{|x-2|}\)
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a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)
= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)
= \(\sqrt{15}.0\)
= 0
b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)
= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)
= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)
= \(\sqrt{5}.\sqrt{5}\)
= 5
c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
= \(\sqrt{5}+\sqrt{3}\)
d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)
= \(\left(2+\sqrt{5}-2+\sqrt{5}\right)\left(2+\sqrt{5}+2-\sqrt{5}\right)\)
= \(2\sqrt{5}.4\)
= \(8\sqrt{5}\)
e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)
= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)
= \(\sqrt{3}.\frac{20}{3}\)
= \(\frac{20\sqrt{3}}{3}\)
a, 320+160−2115
b, (15−3+15+3).5
c, (53+35):15
d, (2+5)2−(2+5)2
e, 1348+375−27−10113
A= ( \(\sqrt{1}\)+\(\sqrt{2}\)+\(\sqrt{3}\) ) + (\(\sqrt{20}\) + \(\sqrt{40}\) + \(\sqrt{60}\))
= (1+1,4+1,7)+(4,4+6,3+7,7)
= 4,1+18,4
=22,5
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)