Rút gọn biểu thức:
a, A= \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
b, \(\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\)
Giúp mih nhanh với ạ. Cần gấp! Cảm ơn nhiều ạ
K
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8 tháng 11 2021
\(a,=\left|2-\sqrt{3}\right|=2-\sqrt{3}\\ b,=\left|3-\sqrt{11}\right|=\sqrt{11}-3\\ c,=2\left|a\right|=2a\\ d,=3\left|a-2\right|=3\left(2-a\right)\left(a< 0\Leftrightarrow a-2< 0\right)\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
3 tháng 2 2022
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
LT
25 tháng 6 2021
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
25 tháng 6 2021
Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
a) A= \(\sqrt{2-\sqrt{3}}\) \(\left(\sqrt{6}-\sqrt{2}\right)\)\(\left(2+\sqrt{3}\right)\)
A= \(\sqrt{2-\sqrt{3}}\) . \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}\) .\(\left(\sqrt{6}-\sqrt{2}\right)\)
A= \(\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\) . \(\sqrt{2+\sqrt{3}}\) . \(\sqrt{2}\left(\sqrt{3}-1\right)\)
A= 1. \(\sqrt{2\left(2+\sqrt{3}\right)}\) \(\left(\sqrt{3}-1\right)\)
A=\(\sqrt{4+2\sqrt{3}}\) .\(\left(\sqrt{3}-1\right)\)
A=\(\sqrt{\left(\sqrt{3}+1\right)^2}\) \(\left(\sqrt{3}-1\right)\)
A=\(\left|\sqrt{3}+1\right|\)\(\left(\sqrt{3}-1\right)\)
A=\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
A=3-1
A=2
Vậy A=2
b)\(\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{\sqrt{2}+\sqrt{3}}\)=\(\frac{\sqrt{2+\sqrt{3}}.1}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) .