B1, Tính a, \(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
b, \(\sqrt{16}.\sqrt{25}+\sqrt{256}.\sqrt{64}\)
c,\(\sqrt{\left(\sqrt{2}-3\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)
K
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TT
20 tháng 8 2017
a. bấm máy tính nó nói max nặng
b.\(\sqrt{16}.\sqrt{25}+\sqrt{256.\sqrt{64}}\)
=\(\sqrt{4^2}.\sqrt{5^2}+\sqrt{16^2.8}\)
=/4/./5/+/16/./2/\(\sqrt{2}\)
=4.5+16.2\(\sqrt{2}\)
=20+32\(\sqrt{2}\)
c.\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)
=/\(2-\sqrt{3}\)/-/5\(-\sqrt{2}\)/
=\(2-\sqrt{3}-5-\sqrt{2}\)
=\(\left(2-5\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
=\(-3-\sqrt{3}-\sqrt{2}\)
= -6,146
8 tháng 7 2021
a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)
b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)
c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)
d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)
30 tháng 7 2021
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: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
AH
Akai Haruma
Giáo viên
30 tháng 9 2023
a.
$A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}$
$A\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}$
$A\sqrt{2}=\sqrt{(\sqrt{3}-1)^2}+\sqrt{(\sqrt{3}+1)^2}$
$=|\sqrt{3}-1|+|\sqrt{3}+1|=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}$
$\Rightarrow A=2\sqrt{3}: \sqrt{2}=\sqrt{6}$
---------------------
$B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$B\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$
$B\sqrt{2}=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$
$=|\sqrt{7}-1|-|\sqrt{7}+1|=\sqrt{7}-1-(\sqrt{7}+1)=-2$
$\Rightarrow B=-2:\sqrt{2}=-\sqrt{2}$
H
30 tháng 9 2023
\(a,\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(A-\sqrt{2}=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)\cdot\sqrt{2}\\ =\sqrt{2-\sqrt{3}}\cdot\sqrt{2}-\sqrt{2+\sqrt{3}}\cdot\sqrt{2}\\ =\sqrt{\left(2-\sqrt{3}\right)\cdot2}-\sqrt{\left(2+\sqrt{3}\right)\cdot2}\\ =\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1-\sqrt{3}-1\\ =-2\)
Ta có :
\(A-\sqrt{2}=-2\\ \Leftrightarrow A=\dfrac{-2}{\sqrt{2}}=\dfrac{-\left(\sqrt{2}\right)^2}{\sqrt{2}}=-\sqrt{2}\)
__
C làm giống câu a, nhé.
__
\(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|2\sqrt{5}+1\right|-\left|\sqrt{5}-2\right|\\ =2\sqrt{5}+1-\sqrt{5}+2\\ =3+\sqrt{5}\)
__
\(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\\ =\sqrt{48-2\cdot4\cdot\sqrt{3}\cdot2+4}+\left|4\sqrt{3}-7\right|\\ =\sqrt{\left(4\sqrt{3}\right)^2-2\cdot4\cdot\sqrt{3}\cdot2+2^2}+4\sqrt{3}-7\\ =\sqrt{\left(4\sqrt{3}-2\right)^2}+4\sqrt{3}-7\\ =4\sqrt{3}-2+4\sqrt{3}-7\\ =8\sqrt{3}-9\)
a,\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=|^{ }_{ }2-\sqrt{5}|^{ }_{ }-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)(vì \(2-\sqrt{5}< 0\))
=-2
b,\(\sqrt{16}\cdot\sqrt{25}+\sqrt{256}\cdot\sqrt{64}\)
\(=4\cdot5-16\cdot8=20+128=148\)
c,\(\sqrt{\left(\sqrt{2}-3\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)
\(=|^{ }_{ }\sqrt{2}-3|^{ }_{ }-|^{ }_{ }5-\sqrt{2}|^{ }_{ }\)
\(=3-\sqrt{2}-5+\sqrt{2}\)(vì \(\sqrt{2}-3< 0;5-\sqrt{2}>0\))
\(=-2\)
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