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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)
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)
\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{2\sqrt{x}+1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}-1-2\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{-2}{2\sqrt{x}-1}\)
Tick hộ nha
a: Ta có: \(A=\left(\sqrt{48}-2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-2\sqrt{45}-\sqrt{3}\)
\(=\left(2\sqrt{3}+2\sqrt{5}\right)\cdot\sqrt{5}-6\sqrt{5}-\sqrt{3}\)
\(=2\sqrt{15}+10-6\sqrt{5}-\sqrt{3}\)
b: Ta có: \(B=\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5}+\sqrt{2}}\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)
\(=\dfrac{2\sqrt{2}}{9+6\sqrt{2}}=\dfrac{-8+6\sqrt{2}}{3}\)
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(x-1\right)^2}{4x}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
b.
\(\left|x-5\right|=4\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{9}+1}{2\sqrt{9}}=\dfrac{2}{3}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
a: \(A=\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}\)
\(=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}\)
\(=3\sqrt{x+5}\)
b: A=6
=>\(3\sqrt{x+5}=6\)
=>\(\sqrt{x+5}=2\)
=>x+5=4
=>x=-1
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
\(\sqrt{\left(8-4x\right)^2}=2\)
\(\Leftrightarrow4x-8=2\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)