\(\frac{1}{6x-4y}-\frac{1}{6x+4y}-\frac{3x}{4y^2-9x^2}\)...">
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\(\dfrac{1}{6x-4y}-\dfrac{1}{6x+4y}-\dfrac{3x}{4y^2-9x^2}\)

\(=\dfrac{1}{2\left(3x-2y\right)}-\dfrac{1}{2\left(3x+2y\right)}+\dfrac{3x}{\left(3x-2y\right)\left(3x+2y\right)}\)

\(=\dfrac{3x+2y-\left(3x-2y\right)+6x}{2\left(3x-2y\right)\left(3x+2y\right)}\)

\(=\dfrac{9x+2y-3x+2y}{2\left(3x-2y\right)\left(3x+2y\right)}=\dfrac{6x+4y}{2\left(3x+2y\right)\left(3x-2y\right)}=\dfrac{1}{3x-2y}\)

6x−4y1−6x+4y1−4y2−9x23x

\(= \frac{1}{2 \left(\right. 3 x - 2 y \left.\right)} - \frac{1}{2 \left(\right. 3 x + 2 y \left.\right)} + \frac{3 x}{\left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)}\)

\(= \frac{3 x + 2 y - \left(\right. 3 x - 2 y \left.\right) + 6 x}{2 \left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)}\)

\(= \frac{9 x + 2 y - 3 x + 2 y}{2 \left(\right. 3 x - 2 y \left.\right) \left(\right. 3 x + 2 y \left.\right)} = \frac{6 x + 4 y}{2 \left(\right. 3 x + 2 y \left.\right) \left(\right. 3 x - 2 y \left.\right)} = \frac{1}{3 x - 2 y}\)

a)có khả năng sai đề bài

b)Liệu có sai đề bài không

c)\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)(phân số cuối có âm vì (1-x)=-(x-1)

\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)(Hơi tắt)

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

d)\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy+4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x^2+4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x-2y}\)

6 tháng 7 2017

a ) \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}=\frac{4\left(x-2\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{6-5x}{\left(x+2\right)\left(x-2\right)}=\frac{6x-4+6-5x}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x+2}\)

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

\(=\frac{-6x^2+5x-1+6x^2-4x+2-3x}{2x\left(2x-1\right)}=\frac{-2x+1}{2x\left(2x-1\right)}=\frac{-1}{2x}\)

c ) \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}=\frac{1}{\left(x+3\right)^2}+\frac{1}{-\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{-12x+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}=\frac{x^3-21x}{x^4-18x^2+81}\)

d ) \(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}=\frac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{x^3-1}=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{1}{x^2+x+1}\)

e ) \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x}{x+2y}\)

17 tháng 12 2014

Cac ban bam' vao` xem them se~ thay' bai` cua minh` 

24 tháng 5 2017

a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)

\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)

\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)

\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)

b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)

\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)

\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)

\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)

\(\Leftrightarrow\frac{1}{3x^2-y^2}\)

17 tháng 12 2019

a) \(\frac{2x-7}{10x-4}-\frac{3x+5}{4-10x}\)

\(=\frac{2x-7}{10x-4}-\frac{-\left(3x+5\right)}{-\left(4-10x\right)}\)

\(=\frac{2x-7}{10x-4}-\frac{5-3x}{10x-4}\)

\(=\frac{2x-7-\left(5-3x\right)}{10x-4}\)

\(=\frac{2x-7-5+3x}{10x-4}\)

\(=\frac{5x-12}{10x-4}\)

31 tháng 3 2020

Làmmmm

1/ \(\frac{1-2x}{2x}+\frac{2x}{2x-1}+\frac{1}{2x-4x^2}\)(ĐKXĐ:x\(\ne0\), x\(\ne\frac{1}{2}\))

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

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

\(=\frac{4x-2}{2x\left(2x-1\right)}=\frac{2\left(2x-1\right)}{2x\left(2x-1\right)}=\frac{1}{x}\)

KL:..............

31 tháng 3 2020

2/\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)(ĐKXĐ : x\(\ne1\))

\(=\frac{x^2+2}{x^3-1}+\frac{2x-2}{x^3-1}-\frac{x^2+x+1}{x^3-1}\)

\(=\frac{x^2+2+2x-2-x^2-x-1}{x^3-1}=\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}\)

Kl:....................