A=(1/x-2 -1/x+2).x+2/2 với x khác +-2
a) Rút gọn A
b) Tìm x để A=2
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12 tháng 9 2021
\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=t\)
\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)
12 tháng 9 2021
a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)
\(=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2+x-2}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
20 tháng 5 2022
\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)
\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)
D
A= (2/x-√x - 1/√x-1) : x-4/x√x+√x - 2x với x>0, x khác 1, x khác 4 a) rút gọn A b) tìm x để A > -1/2
1
5 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{2}{x-\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x-4}{x\sqrt{x}+\sqrt{x}-2x}\)
\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{x-4}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\sqrt{x}+1}{\sqrt{x}+2}\)
17 tháng 7 2023
a) A=2x2+6x-2x2+3x-4x+6+x-2=6x+4
b) x+1=2 => x=1
Tại x=1, A=6*1+4=10
c) A=6x+4=1/2 => x=(1/2-4)/6=-7/12
H
17 tháng 7 2023
`!`
`a,A=2x(x+3) -(x+2)(2x-3)+x-2`
`= 2x^2 + 6x-(2x^2 -3x+4x-6)+x-2`
`= 2x^2 +6x+2x^2 +3x-4x+6+x-2`
`= (2x^2-2x^2)+(6x+3x-4x+x)+(6-2)`
`=6x+4`
`b, x+1=2`
`=>x=2-1`
`=>x=1`
`A=6x+4` mà `x=1`
Thì `6x+4=6.1+4=10`
`c,` Ta có :
`6x+4=1/2`
`=> 6x=1/2-4`
`=> 6x= -7/2`
`=>x=-7/2 : 6`
`=>x=-7/2 xx1/6`
`=>x= -7/12`
19 tháng 1 2022
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
31 tháng 3 2023
a: \(B=\dfrac{\sqrt{x}+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{x+2\sqrt{x}}{2\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: B>2A
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}>2\)
=>-căn x+1>0
=>-căn x>-1
=>căn x<1
=>0<x<1
4 tháng 10 2021
a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
3 tháng 10 2021
Câu a đã làm: F=(2√x/2√x-1 - 1/√x) ( √x+1/√x-1 + 3x/x-2√x+1) với x >0, x khác 1, x khác 1/4 a) rút gọn F - Hoc24
\(b,F=2\Leftrightarrow\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow2\sqrt{x}\left(x-2\sqrt{x}+1\right)=2x\sqrt{x}-4x+2\sqrt{x}+2x-2\sqrt{x}+1\\ \Leftrightarrow2x\sqrt{x}-4x+2\sqrt{x}=2x\sqrt{x}-2x+1\\ \Leftrightarrow2x-2\sqrt{x}+1=0\\ \Leftrightarrow2\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\\ \Leftrightarrow x\in\varnothing\)
a: \(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}=\dfrac{2}{x-2}\)
b: Để A=2 thì 2/x-2=2
=>x-2=1
=>x=3
Tks bạn nhiều!!!