Rút gọn biểu thức :
\(A=5\sqrt{32}-3\sqrt{72}-\sqrt{12,5}\)
\(B=\sqrt{\frac{10+6\sqrt{3}}{1+\sqrt{3}}}\)
K
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27 tháng 8 2023
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
14 tháng 10 2020
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
22 tháng 11 2021
\(a,=4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\\ b,=\dfrac{3\left(\sqrt{2}+1\right)}{1}+\left|3-\sqrt{2}\right|-2\sqrt{2}\\ =3\sqrt{2}+3+3-\sqrt{2}-2\sqrt{2}=6\)
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
24 tháng 8 2023
a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)
\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)
\(=2\left(3-2\sqrt{2}\right)=6-4\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
AH
Akai Haruma
Giáo viên
5 tháng 11 2023
Vì đây toàn là số cụ thể rồi nên không có đkxđ bạn nhé.
Lời giải:
a.
$=\sqrt{2}+4\sqrt{2}+6\sqrt{2}-3\sqrt{2}=8\sqrt{2}$
b.
$=\frac{13(5-2\sqrt{3})}{(5+2\sqrt{3})(5-2\sqrt{3})}+2\sqrt{3}=\frac{13(5-2\sqrt{3})}{13}+2\sqrt{3}$
$=5-2\sqrt{3}+2\sqrt{3}=5$
c.
$=2\sqrt{5}-|2-\sqrt{5}|=2\sqrt{5}-(\sqrt{5}-2)=\sqrt{5}+2$