Tìm x
\(\frac{2x+1}{2}=\frac{4x-1}{3}\)
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Những câu hỏi liên quan
1 tháng 6 2019
\(1,\)\(\frac{x+2}{x+3}+\frac{x-1}{x+1}=\frac{2}{x^2+4x+3}+1\)
\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)
\(\Rightarrow x^2-3x-6=0\)
.....
1 tháng 6 2019
\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)
\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)
.....
16 tháng 12 2020
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
T
3 tháng 7 2019
#)Giải :
\(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-3-x+\frac{1}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\x=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
3 tháng 7 2019
a) \(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-x=\frac{1}{2}+3\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy...
b) \(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-1=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-x=3-\frac{1}{3}+1\)
\(\Leftrightarrow x=\frac{11}{3}\)
Vậy ...
c) \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\frac{49}{50}=\frac{349}{50}+x\)
\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{49}{50}\)
\(\Leftrightarrow x=\frac{199}{25}\)
Vậy ...
K
3
5 tháng 10 2017
\(\frac{4x+3}{2x+1}=\frac{2x+1}{x+2}\)
\(\Rightarrow\frac{\left(4x+2\right)+1}{2x+1}=\frac{\left(2x+4\right)-3}{x+2}\)
\(\Rightarrow\frac{2.\left(2x+1\right)}{2x+1}+\frac{1}{2x+1}=\frac{2.\left(x+2\right)}{x+2}-\frac{3}{x+2}\)
\(\Rightarrow2+\frac{1}{2x+1}=2-\frac{3}{x+2}\)
\(\Rightarrow\frac{1}{2x+1}=\frac{3}{x+2}\)
\(\Rightarrow x+2=3.\left(2x+1\right)\)
\(\Rightarrow x+2=6x+3\)
\(\Rightarrow x-6x=3-2\)
\(\Rightarrow-5.x=1\Rightarrow x=-\frac{1}{5}\)
5 tháng 10 2017
Nhầm 1 chút nha bạn.
\(\Rightarrow\frac{1}{2x+1}=-\frac{3}{x+2}\)
\(\Rightarrow x+2=-3.\left(2x+1\right)\)
\(\Rightarrow x+2=-6x-3\)
\(\Rightarrow6x+x=-3-2\)
\(\Rightarrow7x=-5\Rightarrow x=-\frac{5}{7}\)
S
14 tháng 12 2018
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)
b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep
c, tt
d, cx r
14 tháng 12 2018
a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)
\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)
\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)
\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)
`(2x+1)/2 = (4x -1)/3`
`=> (2x+1) xx3 = (4x -1) xx2`
`=> 6x + 3 = 8x - 2`
`=> 6x - 8x = -3 -2`
`=> -2x = -5`
`=> x = (-5) : (-2)`
`=> x =5/2`
Vây `x =5/2`
(2x+1)/2=(4x−1)/3
\(= > \left(\right. 2 x + 1 \left.\right) x x 3 = \left(\right. 4 x - 1 \left.\right) x x 2\)
\(6x+3=8x-2\)
\(6x-8x=-3-2\)
\(-2x=-5\)
\(x=\left(\right.-5\left.\right):\left(\right.-2\left.\right)\)
\(x=5/2\)
Vây \(x = 5 / 2\)