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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: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
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
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
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: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=-4\sqrt{5}+15\sqrt{2}\)
b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)
\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)
\(=8\sqrt{3}+2\sqrt{2}-4\)
c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3
=6
d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
a: Ta có: \(A=\left(\sqrt{48}-2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-2\sqrt{45}-\sqrt{3}\)
\(=\left(2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-6\sqrt{5}-\sqrt{3}\)
\(=2\sqrt{15}+10-6\sqrt{5}-\sqrt{3}\)
b: Ta có: \(B=\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5}+\sqrt{2}}\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{2}}{9+6\sqrt{2}}=\dfrac{-8+6\sqrt{2}}{3}\)
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
= \(2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
= \(-\sqrt{5}+15\sqrt{2}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
= \(2.7-2\sqrt{21}+7+2\sqrt{21}=14+7=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
= \(6+2\sqrt{6}.\sqrt{5}+5-2\sqrt{30}\)
= \(11+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
= \(\left(\dfrac{1}{2}-\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
= \(4-4\sqrt{2}-12\sqrt{2}+64\sqrt{2}=4+48\sqrt{2}\)
Bài này dễ ẹc ( đâu có khó đâu :)) )
a) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=\sqrt{2^2.5}-\sqrt{3^2.5}+3\sqrt{3^2.2}+\sqrt{6^2.2}\)
\(=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=\left(2-3\right)\sqrt{5}+\left(9+6\right)\sqrt{2}\)
\(=15\sqrt{2}-\sqrt{5}\)
b) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{2^2.7}.\sqrt{7}-2\sqrt{3}.\sqrt{7}+\sqrt{7}.\sqrt{7}+\sqrt{2^2.21}\)
\(=2.7-2\sqrt{21}+7+2\sqrt{21}\)
\(=14+7+\left(2-2\right)\sqrt{21}=21\)
c) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
\(=6+2\sqrt{30}+5-\sqrt{2^2.30}\)
\(=6+5+2\sqrt{30}-2\sqrt{30}=11\)
d) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{2^2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{10^2.2}\right):\dfrac{1}{8}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right).8\)
\(=2\sqrt{2}-12\sqrt{2}+64\sqrt{2}=54\sqrt{2}\)
Hok tốt
AB=AC \(\Rightarrow\Delta ABC\) cân tại A
\(\Rightarrow AH\) đồng thời là phân giác và trung tuyến
\(\Rightarrow\left\{{}\begin{matrix}\widehat{BAH}=\dfrac{1}{2}\widehat{A}=60^0\\BH=\dfrac{1}{2}BC=6\end{matrix}\right.\)
Trong tam giác vuông ABH:
\(tan\widehat{BAH}=\dfrac{BH}{AH}\Rightarrow AH=\dfrac{BH}{tan\widehat{BAH}}=\dfrac{6}{tan60^0}=2\sqrt{3}\)
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
1. Rút gọn biểu thức:
a) \(\sqrt{\dfrac{81}{25}.\dfrac{49}{16}.\dfrac{9}{196}}=\sqrt{\dfrac{81}{25}}.\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{9}{4.49}}=\dfrac{9}{5}.\dfrac{7}{4}.\dfrac{3}{2.7}=\dfrac{9.3}{5.4.2}=\dfrac{27}{40}\)
b) \(\sqrt{72}-5\sqrt{2}-\sqrt{49.3}+\sqrt{48}+\sqrt{12}=\)
\(=\sqrt{9.4.2}-5\sqrt{2}-\sqrt{49.3}+\sqrt{16.3}+\sqrt{4.3}\)
\(=3.2\sqrt{2}-5\sqrt{2}-7\sqrt{3}+4\sqrt{3}+2\sqrt{3}\)
\(=6\sqrt{2}-5\sqrt{2}-7\sqrt{3}+4\sqrt{3}+2\sqrt{3}\)
\(=\sqrt{2}-\sqrt{3}\)
c) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\) \(=\left|2-\sqrt{3}\right|+\left|2+\sqrt{3}\right|=2-\sqrt{3}+2+\sqrt{3}=4\)
d) \(\sqrt{5}+\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=\)
\(=\sqrt{5}+\sqrt{4.5}-\sqrt{9.5}+3\sqrt{9.2}+\sqrt{9.4.2}\)
\(=\sqrt{5}+2\sqrt{5}-3\sqrt{5}+3.3\sqrt{2}+3.2\sqrt{2}\)
\(=\sqrt{5}+2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=15\sqrt{2}\)
tks @duy bùi ngọc hà nha