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a: \(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}=12\sqrt{2}\)
b: \(=5\sqrt{7}-4\sqrt{7}+3\sqrt{7}=4\sqrt{7}\)
c: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
d: \(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
e: \(=\sqrt{5}+\dfrac{2}{5}\sqrt{5}+\sqrt{5}=2.4\sqrt{5}\)
f: \(=\dfrac{1}{5}\sqrt{5}+\dfrac{3}{2}\sqrt{2}+\dfrac{5}{2}\sqrt{2}=\dfrac{1}{5}\sqrt{5}+4\sqrt{2}\)
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)
2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)
3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)
4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)
5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
bài 1
biểu thức có nghĩa khi x, y thỏa mãn đồng thời
\(\left\{{}\begin{matrix}x,y\ne0\\\dfrac{y}{x}\ge0\end{matrix}\right.\Rightarrow x.y>0}\)x, y khác 0
x.y>0
3: \(\sqrt{12-3\sqrt{7}}-\sqrt{12-3\sqrt{7}}=0\)
4: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
a) \(2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}=2\sqrt{2^2.5}-\sqrt{5^2.2}+3\sqrt{4^2.5}-\sqrt{8^2.5}\\ =4\sqrt{5}-5\sqrt{2}+12\sqrt{5}-8\sqrt{5}=8\sqrt{5}-5\sqrt{2}\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=\sqrt{4^2.2}-\sqrt{5^2.2}+\sqrt{3^2.2}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
c) \(3\sqrt{3}+4\sqrt{2}-5\sqrt{27}=3\sqrt{3}+4\sqrt{2}-5\sqrt{3^2.3}=3\sqrt{3}+4\sqrt{2}-15\sqrt{3}=4\sqrt{2}-12\sqrt{3}\)
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}\\ =\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{3+1}\right)^2-1^2}\\ =\dfrac{2\sqrt{3}}{\sqrt{3}}=2\)
e)\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=2^2-\left(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)^2=4-\left(\dfrac{9+6\sqrt{3}+3}{3+2\sqrt{3}+1}\right)\\ =4-\left(\dfrac{6\left(2+\sqrt{3}\right)}{2\left(2+\sqrt{3}\right)}\right)=4-3=1\)
b) \(\sqrt{32}-\sqrt{50}+\sqrt{18}=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=\left(4-5+3\right)\sqrt{2}=2\sqrt{2}\)
\(A=2\sqrt{18}-4\sqrt{32}+\sqrt{72}+3\sqrt{8}\\ =2\sqrt{18}-16\sqrt{2}+2\sqrt{18}+6\sqrt{2}\\ =4\sqrt{18}-10\sqrt{2}\\ =12\sqrt{2}-10\sqrt{2}=2\sqrt{2}\)
\(B=\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2-\sqrt{3}+2}{3-4}=\dfrac{4}{-1}=-4\)
câu g
(câu cuối) đề nhiều trôi hết nhìn thấy mỗi câu (g)
\(G=0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(G=0,1.10\sqrt{2}+\dfrac{2.2}{10}\sqrt{2}+0,4.5\sqrt{2}\)
\(G=\sqrt{2}\left(1+\dfrac{2}{5}+2\right)=\dfrac{\sqrt{2}\left(5+2+10\right)}{5}=\dfrac{17\sqrt{2}}{5}\)
a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)
=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)
= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)
= \(24\sqrt{2}\)
b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)
= \(\sqrt{7+2\sqrt{7}+1}+\sqrt{7-4\sqrt{7}+4}\)
= \(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-2\right)^2}\)
= \(\sqrt{7}+1+\sqrt{7}-2\)
= \(2\sqrt{7}-1\)
c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
= \(2\sqrt{5}+6-2\sqrt{5}-3\)
= 3
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
\(=\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left(3+2\sqrt{2}-\left(3+2\sqrt{2}\right)\right)\)
\(=\left(9\sqrt{2}\right)+\left(3-2\sqrt{2}-3-2\sqrt{2}\right)\)
\(=9\sqrt{2}+\left(-4\sqrt{2}\right)\)
\(=9\sqrt{2}-4\sqrt{2}\)
\(=5\sqrt{2}\)
\(\left(\sqrt{8}-2\sqrt{32}+3\sqrt{50}\right)+\left(\dfrac{1}{3+2\sqrt{2}}-\dfrac{1}{3-2\sqrt{2}}\right)\)
= \(\left(2\sqrt{2}-8\sqrt{2}+15\sqrt{2}\right)+\left(\dfrac{3-2\sqrt{2}-3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right)\)
= \(9\sqrt{2}+\left(\dfrac{-4\sqrt{2}}{1}\right)\) = \(9\sqrt{2}-4\sqrt{2}\) = \(5\sqrt{2}\)
a, \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{5}+\dfrac{1}{2}.2\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
b, \(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)
\(=\sqrt{0,5}+3\sqrt{0,5}+5\sqrt{0,5}=9\sqrt{0,5}\)
c, \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-3\sqrt{5}+3\sqrt{18}+2\sqrt{18}\)
\(=-\sqrt{5}+5\sqrt{18}\)
d, \(0,1.\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(=\sqrt{0,01.200}+0,2.\sqrt{2}+0,4.5\sqrt{2}\)
\(=\sqrt{2}+0,2\sqrt{2}+2\sqrt{2}=3,2\sqrt{2}\)
Chúc bạn học tốt!!!
\(=5\sqrt{2}-\dfrac{3}{2}\cdot4\sqrt{2}-\dfrac{1}{3}\cdot6\sqrt{2}+8=-3\sqrt{2}+8\)
mình cam ơn nha