rút gọn biểu thức: \(2\sqrt{20}-\sqrt{50}-3\sqrt{80}-\sqrt{320}\)
Thực hiện phép tính (rút gọn biểu thức)
a)\(\sqrt{20}\)-3\(\sqrt{45}\)-\(\dfrac{1}{2}\sqrt{80}\)
b) 12\(\sqrt{54}\)-\(\dfrac{2}{5}\)\(\sqrt{150}\)+3\(\sqrt{24}\)
Lời giải:
a.
$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$
b.
$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$
rút gọn biểu thức : \(\dfrac{5-2\sqrt{5}}{\sqrt{5}}\) - ( \(5\sqrt{5}-3\) ) + \(\sqrt{80}\)
\(\dfrac{5-2\sqrt{5}}{\sqrt{5}}-\left(5\sqrt{5}-3\right)+\sqrt{80}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}-5\sqrt{5}+3+4\sqrt{5}\\ =\sqrt{5}-2-5\sqrt{5}+3+4\sqrt{5}\\ =\sqrt{5}\left(1-5+4\right)-2+3\\ =0+1\\ =1\)
\(=\sqrt{5}-2-5\sqrt{5}+3+4\sqrt{5}=1\)
Rút gọn các biểu thức sau:
\(a,\left(\sqrt{45}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
\(b,\left(\sqrt{5}-\sqrt{3}\right)^2+2\sqrt{15}\)
Bài 1
a. Tìm điều kiện để căn thức bậc hai có nghĩa \(\sqrt{\dfrac{2x+1}{x^2+1}}\)
b. \(\sqrt[3]{-27}+\sqrt[3]{64}-\dfrac{\sqrt[3]{-128}}{\sqrt[3]{2}}\)
* Rút gọn biểu thức
a. \(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)
b. \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
c. \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
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 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
Rút gọn biểu thức:
\(M=\sqrt[3]{26+15\sqrt{3}}.\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(=2\sqrt{4.5}+\sqrt{2.25}+3\sqrt{16.5}-\sqrt{64.5}\)
\(=2.2\sqrt{5}+5\sqrt{2}+3.4\sqrt{5}-8\sqrt{5}\)
\(=\left(4+12-8\right)\sqrt{5}+5\sqrt{2}\)
\(=8\sqrt{5}+5\sqrt{2}\)
P/s: Em mới lớp 5 nên làm đại, sai thì thông cảm ạ.
\(2\sqrt{20}+\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(=2\sqrt{4.5}+\sqrt{25.2}+3\sqrt{16.5}-\sqrt{64.5}\)
\(=2.2\sqrt{5}+3.4\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)
\(=4\sqrt{5}+12\sqrt{5}-8\sqrt{5}+5\sqrt{2}\)
\(=8\sqrt{5}+5\sqrt{2}\)
Rút gọn biểu thức
a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)
`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
\(\sqrt{32}\)+\(\sqrt{50}\) - 2\(\sqrt{200}\) + 3\(\sqrt{72}\)
b)\(\dfrac{3}{\sqrt{ }2-1}\) + \(\sqrt{\left(3-\sqrt{2}\right)^{^2}}\) - 2\(\sqrt{2}\)
rút gọn các biểu thức trên
\(a.4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\)
\(a,=4\sqrt{2}+5\sqrt{2}-20\sqrt{2}+18\sqrt{2}=7\sqrt{2}\\ b,=\dfrac{3\left(\sqrt{2}+1\right)}{1}+\left|3-\sqrt{2}\right|-2\sqrt{2}\\ =3\sqrt{2}+3+3-\sqrt{2}-2\sqrt{2}=6\)
`a)`
`\sqrt{32} + \sqrt{50} - 2\sqrt{200} + 3\sqrt{72}`
`= 4\sqrt{2} + 5\sqrt{2} - 20\sqrt{2} + 18\sqrt{2}`
`= (4 + 5 - 20 + 18) . \sqrt{2}`
`= 7\sqrt{2}`
`b)`
`3/(\sqrt{2} - 1) + \sqrt{(3 - \sqrt{2})^2} - 2\sqrt{2}`
`= (3 . (\sqrt{2} + 1))/1 + |3 - \sqrt{2}| - 2\sqrt{2}`
`= 3\sqrt{2} + 3 + 3 - \sqrt{2} - 2\sqrt{2}`
`= (3 - 1 - 2) . \sqrt{2} + 6`
`= 6`
rút gọn biểu thức
a) \(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{9+2\sqrt{14}}\right)\)
b) \(\sqrt{\sqrt{13}-\sqrt{3-\sqrt{13}}-4\sqrt{3}}\)
c) \(\sqrt{80-\sqrt{321-16\sqrt{5}}-\sqrt{226-80\sqrt{5}-\sqrt{89-25\sqrt{5}}}}\)
d) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
e) \(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
f) \(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3}+\sqrt{5}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
g) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
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: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)