A=√20-√45+3√18+√72
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Tính 1 phần 3 căn 18 + 2 căn 45 - căn 72 + 3 căn 20
\(\dfrac{1}{3}\sqrt{18}+2\sqrt{45}-\sqrt{72}+3\sqrt{20}\)
\(=\dfrac{1}{3}\sqrt{9\times2}+2\sqrt{9\times5}-\sqrt{9\times8}+3\sqrt{4\times5}\)
\(=\sqrt{2}+6\sqrt{5}-6\sqrt{2}+6\sqrt{5}\) \(=12\sqrt{5}-5\sqrt{2}\)
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\(\dfrac{1}{3}\sqrt{18}+2\sqrt{45}-\sqrt{72}+3\sqrt{20}\)
\(=\sqrt{2}+6\sqrt{5}-6\sqrt{2}+6\sqrt{5}\)
\(=12\sqrt{5}-5\sqrt{2}\)
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a,√55.77.35 b,√132-122 c,√20 - √45 + 5√18 + √72 giúp mik với ạ
a: \(=\sqrt{5\cdot7\cdot11\cdot11\cdot35}=\sqrt{35^2\cdot11^2}=35\cdot11=385\)
b: \(=\sqrt{169-144}=\sqrt{25}=5\)
c: \(=2\sqrt{5}-3\sqrt{5}+5\cdot3\sqrt{2}+6\sqrt{2}\)
\(=-\sqrt{5}+21\sqrt{2}\)
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1. Tính:
√20-√45+3√18+√72.
√20-√45+3√18+√72
=2√5-3√5+9√2+6√2
=-√5+15√2
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a,tính : (√12+2√27-3√3)√3
b,tính :√20-√45+3√18+√72
c,tìm x biết :√(2x-1)2 bình phương =3
rút gọn biểu thức sau:\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{4.5}-\sqrt{9.5}+3\sqrt{18}+\sqrt{4.18}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
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\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
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tính bằng cách hợp lí
a ) 27 + 45 + 18 + 81
b) ( 24 + 72 + 36 + 60 ) : 20
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
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e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
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a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
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Rút gọn hai biểu thức sau:
a) \(\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)
b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
a) \(\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\sqrt{\frac{1}{2}}+\sqrt{\frac{9}{2}}+\sqrt{\frac{25}{2}}=\sqrt{\frac{1}{2}}+3\sqrt{\frac{1}{2}}+5\sqrt{\frac{1}{2}}=9\sqrt{\frac{1}{2}}\)
b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\sqrt{4.5}-\sqrt{9.5}+3\sqrt{9.2}+\sqrt{36.2}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
a) \(\sqrt{\frac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}\)
b) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}=15\sqrt{2}-\sqrt{5}\)