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12 tháng 1 2017

a) ( 75  - 3 2  -  12 )( 3  +  2 )

=(5 3 - 3 2 - 2 3 )( 3  +  2 )

=3( 3  -  2 )( 3  +  2 ) = 3

19 tháng 11 2023

a: Sửa đề: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\cdot\left(\sqrt{3}-1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}-1\right)^2\cdot\left(\sqrt{3}-1\right)}\)

\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}=\sqrt{3}-1\)

b: \(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(\sqrt{3}\right)^3+3\cdot\left(\sqrt{3}\right)^2\cdot1+3\cdot\sqrt{3}\cdot1^2+1^3}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(\sqrt{3}+1\right)^3}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

7 tháng 7 2019

a) \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)

\(=\left(1+\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2\)

\(=1+2\sqrt{2}+2-3\)

\(=2\sqrt{2}\)

7 tháng 7 2019

b) \(\left(1+2\sqrt{3}-\sqrt{2}\right)\left(1+2\sqrt{3}+\sqrt{2}\right)\)

\(=\left(1+2\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)

\(=1+4\sqrt{3}+12-2\)

\(=9+4\sqrt{3}\)

5 tháng 7 2023

\(\dfrac{2}{\sqrt{3}-1}-\dfrac{3\sqrt{3}}{\sqrt{3}+1}\\ =\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{3\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\left(\sqrt{3}+1\right)-3\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\\ =\dfrac{2\sqrt{3}+2-9+3\sqrt{3}}{2}\\ =\dfrac{5\sqrt{3}-7}{2}\)

8 tháng 11 2021

\(=\dfrac{2\sqrt{3}+2+2\sqrt{3}-2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\dfrac{4\sqrt{3}}{2}-\sqrt{3}=2\sqrt{3}-\sqrt{3}=\sqrt{3}\)

12 tháng 9 2023

a) \(\left(2\sqrt{2}-3\right)^2\)

\(=\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot3+3^2\)

\(=4\cdot2-12\sqrt{2}+9\)

\(=17-12\sqrt{2}\)

b) \(\sqrt{\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)^2}\)

\(=\left|\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right|\)

\(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}-1}{2}\)

c) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)

\(=\left|0,1-\sqrt{0,1}\right|\)

\(=0,1-\sqrt{0,1}\)

16 tháng 10 2021

a) \(P=\dfrac{\sqrt{3}+\sqrt{6}}{1+\sqrt{2}}=\dfrac{\left(\sqrt{3}+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}\)

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

16 tháng 10 2021

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

1 tháng 8 2021

Q=\(\dfrac{\left(3+2\sqrt{3}\right)\left(\sqrt{2}+1\right)+\sqrt{3}\left(2+\sqrt{2}\right)}{\sqrt{3}\left(\sqrt{2}+1\right)}-\left(\sqrt{2}+\sqrt{3}\right)\)

Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)-\(\left(\sqrt{2}+\sqrt{3}\right)\)

Q=\(\dfrac{3+4\sqrt{3}+3\sqrt{6}+3\sqrt{2}-2\sqrt{3}-3\sqrt{2}-\sqrt{6}-3}{\sqrt{3}\left(\sqrt{2}+1\right)}\)

Q=\(\dfrac{2\sqrt{3}-2\sqrt{6}}{\sqrt{3}\left(\sqrt{2}+1\right)}\)=\(\dfrac{\sqrt{4}-\sqrt{8}}{\sqrt{2}+1}\)

3 tháng 1 2021

1.

\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)

2. 

a, ĐK: \(x\in R\)

\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)

\(\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, ĐK: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)