\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+\sqrt{98}\right)\)
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Tính:
\(\left(2\sqrt{8}-\sqrt{45}-\sqrt{98}\right)\left(\sqrt{42}+\sqrt{20}-\sqrt{32}\right)\)
\(=\left(4\sqrt{2}+3\sqrt{2}-7\sqrt{2}\right)\left(\sqrt{42}+2\sqrt{5}-\sqrt{32}\right)=0.\left(\right)=0\)
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6) Thực hiện các phép tính
a. \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
b. \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
c. \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
Mng giúp mik vs ạ >.<!!
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
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a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)
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c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)
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Xem thêm câu trả lời
Tính:Asqrt{20}-10sqrt{dfrac{1}{5}}+sqrt{left(sqrt{5}-1right)^2}B2sqrt{32}+5sqrt{8}-4sqrt{32}Csqrt{left(3-sqrt{2}^2right)}-sqrt{left(1-sqrt{2}right)^2}Dsqrt{left(5-1right)^2}+sqrt{left(sqrt{5}-3right)^2}Eleft(3+dfrac{5-sqrt{5}}{sqrt{5}-1}right)left(3-dfrac{5+sqrt{5}}{sqrt{5}-1}right)Fsqrt{6+2sqrt{5}}-sqrt{9-4sqrt{5}}Gdfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}Hdfrac{10}{sqrt{3}-1}-dfrac{55}{2sqrt{3}+1} help
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Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
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: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
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Rút gọn:a) 2sqrt{98}-3sqrt{18}+dfrac{1}{2}sqrt{32}b)left(5sqrt{2}+2sqrt{5}right).sqrt{5}-sqrt{250}c)left(2sqrt{3}-5sqrt{2}right).sqrt{3}-sqrt{36}d)3sqrt{48}+2sqrt{27}-dfrac{1}{3}sqrt{243}e) 6sqrt{dfrac{1}{3}}+dfrac{9}{sqrt{3}}-dfrac{2}{sqrt{3}-1}f)4sqrt{dfrac{1}{2}}-dfrac{6}{sqrt{2}}dfrac{2}{sqrt{2}+1}
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Rút gọn:
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}\)
b)\(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}\)
c)\(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}\)
d)\(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)
f)\(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}\dfrac{2}{\sqrt{2}+1}\)
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)
c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)
d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)
f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)
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Bài 1:Tính GT các biểu thức:
A= \(\left(\sqrt{5}+\sqrt{3}\right)^2\) - \(\left(\sqrt{5}-\sqrt{8}\right)^2\)
B= \(\sqrt{50}-3\sqrt{98}+2\sqrt{8}-3\sqrt{32}-5\sqrt{18}\)
C= \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
(mink đag cần gấp)
Tính
a)\(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
b)\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) với \(a\ge0\)
c)\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)
d)\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right).\sqrt{11}+3\sqrt{32}\)
Lời giải:
a)
$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$
$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$
b)
$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$
$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$
$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$
c)
$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$
$=6-\sqrt{15}$
d)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$
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A = \(10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)
\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)
\(A=10-\left(27-8\right)\)
\(A=-9\)
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Tính:
\(a)D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\left(-\sqrt{2}\right)\\ b)2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\\ c)E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\\ d)P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(e)M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)
a) \(\left(3-a\right)^2\sqrt{0,2}.\sqrt{180a^2}\)
b) \(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
c) \(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{a-2\sqrt{a+1}}\right)\div\frac{\sqrt{a-1}}{a-2\sqrt{a+}1}\)với \(a>0\)và \(a\ne1\)