b4 : rút gọn
A=\(\sqrt{7-\sqrt{27}}+\sqrt{7+\sqrt{24}}\)
B= \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
K
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H9
HT.Phong (9A5)
CTVHS
14 tháng 7 2023
a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)
\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)
\(=14\sqrt{3}-2\sqrt{57}\)
b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
17 tháng 1 2022
, \(A=\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
\(=\frac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{3}-\frac{5\left(4-\sqrt{7}\right)}{9}\)
\(=\frac{-16+4\sqrt{7}}{4}+\frac{18\sqrt{7}+36-20+5\sqrt{7}}{9}=-4+\sqrt{7}+\frac{23\sqrt{7}+16}{9}\)
b,\(B=\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{2}+\frac{5\sqrt{6}}{6}\)
\(=\frac{12\sqrt{6}+5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
31 tháng 10 2021
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
9 tháng 10 2021
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
24 tháng 8 2023
a: Sửa đề: \(\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{3}-2}=\dfrac{2-\sqrt{3}}{\sqrt{3}-2}\)
=-1
b: Sửa đề: \(\dfrac{\sqrt{5-2\sqrt{6}}}{\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
=1
18 tháng 1 2022
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 \(M=\sqrt[3]{\left(\sqrt{2}+1\right)^3}=\sqrt{2}+1\)
18 tháng 1 2022
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 \(N=\sqrt[3]{\left(\sqrt{3}-1\right)^3=\sqrt{3}-1}\)
MT
2 tháng 10 2015
\(A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}\)
\(=\sqrt{5}-\sqrt{7}-1\)
a: Sửa đề: \(A=\sqrt{7-\sqrt{24}}+\sqrt{7+\sqrt{24}}\)
\(=\sqrt{6}-1+\sqrt{6}+1=2\sqrt{6}\)
b: \(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)