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1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
Ta có: 3 x 2 + 2 5 x - 3 3 = - x 2 - 2 3 x +2 5 +1
⇔ 3 x 2 + 2 5 x - 3 3 + x 2 + 2 3 x - 2 5 – 1= 0
⇔ ( 3 +1) x 2 + (2 5 + 2 3 )x -3 3 - 2 5 – 1= 0
⇔ ( 3 +1)x2 + 2( 5 + 3 )x -3 3 - 2 5 – 1= 0
∆ ' = b ' 2 – ac= 3 + 5 2 – ( 3 + 1 )( -3 3 - 2 5 – 1)
= 5 + 2 15 +3+9 +2 15 + 3 +3 3 +2 5 + 1
=18 +4 15 +4 3 +2 5
= 1 + 12 + 5 + 2.2 3 + 2 5 + 2.2 3 . 5
= 1 + 2 3 2 + 5 2 + 2.1.2 3 +2.1. 5 + 2.2 5 . 3
= 1 + 2 3 + 5 2 > 0
a: \(=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
b: \(=\sqrt{3}-1+2-\sqrt{3}=1\)
c: \(=2-\sqrt{3}+2-\sqrt{3}=4-2\sqrt{3}\)
a: \(=5\cdot5\sqrt{3}-\dfrac{1}{3}\cdot3\sqrt{3}=24\sqrt{3}\)
b: \(=\dfrac{12\left(3+\sqrt{5}\right)}{4}=9+3\sqrt{5}\)
c: \(=3-\sqrt{5}+\sqrt{5}=3\)
a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
`\sqrt{(3-\sqrt{5})^2}+\sqrt{5}=|3-\sqrt{5}|+\sqrt{5}=3-\sqrt{5}+\sqrt{5}=3`
`\sqrt{3}-\sqrt{(1+\sqrt{3})^2}=\sqrt{3}-|1+\sqrt{3}|=\sqrt{3}-1-\sqrt{3}=-1`
`\sqrt{(\sqrt{3}-1)^2}-\sqrt{3}=|\sqrt{3}-1|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1`
\(\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{5}=\left|3-\sqrt{5}\right|+\sqrt{5}=3-\sqrt{5}+\sqrt{5}=3\)
\(\sqrt{3}-\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{3}-\left|1+\sqrt{3}\right|=\sqrt{3}-1-\sqrt{3}=-1\)
\(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)