rút gọn biểu thức: \(2\sqrt{20}-\sqrt{50}-3\sqrt{80}-\sqrt{320}\)
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Lời giải:
a.
$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$
b.
$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$
Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(=2\sqrt{4.5}+\sqrt{2.25}+3\sqrt{16.5}-\sqrt{64.5}\)
\(=2.2\sqrt{5}+5\sqrt{2}+3.4\sqrt{5}-8\sqrt{5}\)
\(=\left(4+12-8\right)\sqrt{5}+5\sqrt{2}\)
\(=8\sqrt{5}+5\sqrt{2}\)
P/s: Em mới lớp 5 nên làm đại, sai thì thông cảm ạ.
\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(=2\sqrt{4.5}+\sqrt{25.2}+3\sqrt{16.5}-\sqrt{64.5}\)
\(=2.2\sqrt{5}+3.4\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)
\(=4\sqrt{5}+12\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)
\(=8\sqrt{5}+5\sqrt{2}\)
`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`
`=14-12\sqrt{7}+18+12\sqrt{7}=32`
`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`
`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`
`=3\sqrt{5}`.
a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)
\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)
\(=14-6\sqrt{28}+18+6\sqrt{28}\)
\(=14+18\)
\(=32\)
b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)
\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)
\(=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}\)
\(=3\sqrt{5}\)
\(a,=4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\\ b,=\dfrac{3\left(\sqrt{2}+1\right)}{1}+\left|3-\sqrt{2}\right|-2\sqrt{2}\\ =3\sqrt{2}+3+3-\sqrt{2}-2\sqrt{2}=6\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
\(2\sqrt{20}-\sqrt{50}-3\sqrt{80}-\sqrt{320}=2\sqrt{2^2.5}-\sqrt{2.5^2}-3\sqrt{\left(2^2\right)^2.5}-\sqrt{\left(2^3\right)^2.5}=4\sqrt{5}-5\sqrt{2}-12\sqrt{5}-8\sqrt{5}=\left(4-12-8\right).\sqrt{5}-5\sqrt{2}=-16\sqrt{5}-5\sqrt{2}\)