rút gọn biểu thức
\(\left(\sqrt{45}-\sqrt{63}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
K
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Những câu hỏi liên quan
10 tháng 9 2019
\(\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)\)
\(\Rightarrow5\sqrt{5}-\sqrt{75}+5\sqrt{3}-\sqrt{45}\)
\(\Rightarrow5\sqrt{5}-5\sqrt{3}+5\sqrt{3}-\sqrt{45}\)
\(\Rightarrow5\sqrt{5}-3\sqrt{5}\)
\(\Rightarrow2\sqrt{5}\)
\(\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(\Rightarrow\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(\Rightarrow3\sqrt{35}-15+21-3\sqrt{35}\)
\(\Rightarrow6\)
HM
14 tháng 10 2017
A= \(\sqrt{5}^2-2^2=5-4=1\)
B=(\(\sqrt{9}\cdot\sqrt{5}+\sqrt{9}\cdot\sqrt{7}\) )(\(\sqrt{7}-\sqrt{5}\))
=\(\sqrt{9}\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=\(\sqrt{9}\left(\sqrt{7}^2-\sqrt{5}^2\right)\)
=\(\sqrt{3^2}\cdot\left(7-5\right)=3\cdot2=6\)
C=\(\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}\left(\sqrt{5}-\sqrt{3}\right)\right)\)
=\(\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=\(\sqrt{5}\left(5-3\right)\)=\(2\sqrt{5}\)
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.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
8 tháng 7 2018
\(\sqrt{3+\sqrt{5}}=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}+1}{\sqrt{2}}\)
\(\sqrt{7+3\sqrt{5}}=\frac{\sqrt{14+2.3\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{9+2.3\sqrt{5}+5}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}=\frac{3+\sqrt{5}}{\sqrt{2}}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{3+2.\sqrt{3}.3\sqrt{2}+18}=\sqrt{\left(\sqrt{3}+3\sqrt{2}\right)^2}=\sqrt{3}+3\sqrt{2}\)
\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=3\sqrt{2}-\sqrt{3}\)
Nên \(E=\frac{\sqrt{5}+1+3+\sqrt{5}}{\sqrt{2}}.\left(3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\right)\)
\(=\frac{4+2\sqrt{5}}{\sqrt{2}}.2.3.\sqrt{2}=24+12\sqrt{5}\)
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: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 4 2021
\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1
\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)
\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)
11 tháng 1
\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)
\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)
\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)
b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)
\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)
\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)
\(\left(\sqrt{45}-\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(\sqrt{9.5}-\sqrt{9.7}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=-3\left(\sqrt{5}-\sqrt{7}\right)^2\)