1 Rút gọn biểu thức
a) căn 75 +căn 48 - căn 300
b) căn 94 - căn 16a + căn 49a ( a>- 0)
c) căn 1666 - 2 căn 406 - 3 căn 906 ( b >- 0 )
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12 tháng 7 2023
\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
27 tháng 12 2017
a)
\(7\sqrt{12}+\frac{1}{3}\sqrt{27}-\sqrt{75}\)
\(=14\sqrt{3}+\sqrt{3}-5\sqrt{3}\)
\(=10\sqrt{3}\)
b)
\(\left(2\sqrt{20}+\sqrt{125}-3\sqrt{80}\right):5\)
\(=\left(4\sqrt{5}+5\sqrt{5}-12\sqrt{5}\right):5\)
\(=-3\sqrt{5}:5\)
\(=\frac{-3\sqrt{5}}{5}\)
c)
\(3\sqrt{12a}-5\sqrt{3a}+\sqrt{48a}\)
\(=6\sqrt{3a}-5\sqrt{3a}+4\sqrt{3a}\)
\(=5\sqrt{3a}\)
15 tháng 11 2022
a: \(=9\sqrt{2}-4\sqrt{2}+4\sqrt{2}+9\sqrt{2}=18\sqrt{2}\)
b: \(=8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
c: \(=2\sqrt{21}\)
PV
1
LS
Lê Song Phương
CTVHS
8 tháng 11 2021
\(\left(\sqrt{75}+\sqrt{243}-\sqrt{48}\right):\sqrt{3}\)
\(=\sqrt{75}:\sqrt{3}+\sqrt{243}:\sqrt{3}-\sqrt{48}:\sqrt{3}\)
\(=\sqrt{75:3}+\sqrt{243:3}-\sqrt{48:3}\)
\(=\sqrt{25}+\sqrt{81}-\sqrt{16}\)
\(=5+9-4=10\)
23 tháng 8 2023
a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)
\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)
\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)
\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)
\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)
\(=-8\sqrt{3}\)
b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)
\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)
19 tháng 2 2021
a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)
\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)(1)
b) Ta có: \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào biểu thức (1), ta được:
\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)
