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16 tháng 12 2016

\(A=\left(\frac{2\left(2x-1\right)+4x^2+1+2x+1}{4x^2-1}\right):\frac{2}{4x^2-1}\)

\(A=\left(\frac{4x^2+6x}{4x^2-1}\right).\frac{4x^2-1}{2}=\left(\frac{2x\left(2x+3\right)}{4x^2-1}\right).\frac{4x^2-1}{2}\)

\(\orbr{\begin{cases}x\ne+-\frac{1}{2}\\A=x\left(2x+3\right);\end{cases}}\)

\(A=2\Rightarrow2x^2+3x-2=0\Rightarrow\orbr{\begin{cases}x=\frac{-3+5}{4}=\frac{1}{2}\left(loai\right)\\x=\frac{-3-5}{4}=-2\end{cases}}\)

1 tháng 8 2015

a/ \(=\left(\frac{2\left(1-2x\right)-\left(4x^2+1\right)-\left(1+2x\right)}{1-4x^2}\right).\frac{4x^2-1}{2}=\frac{2-4x-4x^2-1-1-2x}{1-4x^2}.\frac{4x^2-1}{2}=\frac{-4-6x-4x^2}{1-4x^2}.\frac{4x^2-1}{2}=\frac{4x^2+6x+4}{2}=2x^2+3x+2\)

b/ có A = 2 \(\Leftrightarrow2x^2+3x+2=2\Rightarrow2x^2+3x=0\Rightarrow x\left(2x+3\right)=0\Rightarrow x=0\)

                                                                                                              hoặc \(2x+3=0\Rightarrow2x=-3\Rightarrow x=-\frac{3}{2}\)

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

\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)

\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)

\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)

\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)

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) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)