Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=2\sqrt3-\sqrt{75}+2\sqrt{12}\)
\(=2\sqrt3-5\sqrt3+2\cdot2\sqrt3\)
\(=-3\sqrt3+4\sqrt3=\sqrt3\)
b: \(B=\sqrt{\left(2-\sqrt5\right)^2}+\sqrt{\left(3-\sqrt5\right)^2}\)
\(=\left|2-\sqrt5\right|+\left|3-\sqrt5\right|\)
\(=\sqrt5-2+3-\sqrt5=-2+3=1\)
c: \(C=\left(\frac{x+2\sqrt{x}}{x-2\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\frac{1}{\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\frac{1}{\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}-2}\)
Lời giải:
a. $P=1+\sqrt{(1-\sqrt{2})^2}=1+|1-\sqrt{2}|=1+\sqrt{2}-1=\sqrt{2}$
b.
$\sqrt{x-1}=3$ (đk: $x\geq 1$)
$\Leftrightarrow x-1=3^2=9$
$\Leftrightarrow x=9+1=10$ (thỏa mãn)
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
`1a)A=(5sqrt5+2sqrt{45}+sqrt5).sqrt5`
`=(5sqrt{5}+2sqrt{9.5}+sqrt5).sqrt5`
`=(5sqrt5+6sqrt5+sqrt5).sqrt5`
`=12sqrt5*sqrt5=60`
`b)B=2sqrt2+1/(2sqrt2)-4sqrt{50}`
`=2sqrt2+1/(2sqrt2)-4.sqrt{25.2}`
`=2sqrt2+1/(2sqrt2)-20sqrt2`
`=(8+1-80)/(2sqrt2)`
`=(-71)/(2sqrt2)`
`=(-71sqrt2)/4`
`c)=1/(1-sqrt2)+1/(1+sqrt2)`
`=(sqrt2+1)/(1-2)+(sqrt2-1)/(2-1)`
`=-sqrt2-1+sqrt2-1=-2`
`2)sqrt{(8-4x)^2}=2`
`<=>|8-4x|=2`
`<=>|4-2x|=1`
`<=>` \(\left[ \begin{array}{l}4-2x=1\\4-2x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=3\\2x=5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=\dfrac52\end{array} \right.\)
Vậy `S={3/2,5/2}`
\(A=\left(5\sqrt{5}+6\sqrt{5}+\sqrt{5}\right).\sqrt{5}=12\sqrt{5}.\sqrt{5}=12.5=60\)
\(B=2\sqrt{2}+\dfrac{1}{2}\sqrt{2}-10\sqrt{2}=-7,5.\sqrt{2}\)
\(C=\dfrac{1}{1-\sqrt{2}}+\dfrac{1}{1+\sqrt{2}}=\dfrac{1+\sqrt{2}}{-1}+\dfrac{1-\sqrt{2}}{-1}=-1-\sqrt{2}-1+\sqrt{2}=-2\)
Bài 2:
\(\sqrt{\left(8-4x\right)^2}=2\)
*TH1: x < 2
\(\sqrt{\left(8-4x\right)^2}=2\)
\(\Leftrightarrow8-4x=2\Leftrightarrow4x=6\Leftrightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
*TH2: x ≥ 2
\(\sqrt{\left(8-4x\right)^2}=2\)
\(\Leftrightarrow4x-8=2\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)