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18 tháng 12 2016

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\left(ĐK:x\ne0;x\ne-4\right)\)

\(=\frac{6}{x\left(x+4\right)}+\frac{3}{2\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(4+x\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

b) \(\frac{4xy-5}{140x^3y}-\frac{6y^2-5}{10x^3y}\left(ĐK:x,y\ne0\right)\)

\(=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{2y\left(2x-3y\right)}{10x^3y}=\frac{2x-3y}{5x^3}\)

19 tháng 12 2016

- cảm ơn bạn nhiều nha !

21 tháng 1 2020

\(\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-6y^2}{10x^3y}=\frac{2y\left(2x-3y\right)}{2y.5x^3}=\frac{2x-3y}{5x^3}\)

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x+1\right)\left(2x-1\right)}:\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1+2x-1\right)\left(2x+1-2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\times\frac{10x-5}{4x}\)
\(=\frac{4x.2}{\left(2x+1\right)\left(2x-1\right)}\times\frac{5\left(2x-1\right)}{4x}\)
\(=\frac{10}{2x+1}\)

21 tháng 1 2020

\(a,\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{4xy-6y^2}{10x^3y}\)

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)

\(=\left(\frac{2x+1}{2x-1}+\frac{2x-1}{2x-1}\right):\frac{4x}{10x-5}\)

\(=\frac{2x+1+2x-1}{2x-1}:\frac{4x}{10x-5}\)

\(=\frac{4x}{2x-1}.\frac{10x-5}{4x}\)

\(=\frac{10x-5}{2x-1}\)

\(=\frac{5\left(2x-1\right)}{2x-1}\)

\(=\frac{5}{1}=5\)

23 tháng 8 2016

a) \(\frac{x^2-y^2}{\left(x+y\right)\left(ay-\text{ax}\right)}=\frac{\left(x+y\right)\left(x-y\right)}{-a\left(x+y\right)\left(x-y\right)}=\frac{-1}{a}\)

b) \(\frac{2ax-2x-3y+3ay}{4ax+\text{4x}+6y+6ay}=\frac{2x\left(a-1\right)+3y\left(a-1\right)}{\text{4x}\left(a+1\right)+6y\left(a+1\right)}\)

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

23 tháng 8 2016

lêu lêu lêu 

3 tháng 12 2019

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

\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)

3 tháng 12 2019

c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)

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

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

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

\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)

d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)

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

e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)

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

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

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

\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\) 

f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)

\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)

3 tháng 8 2020

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

3 tháng 8 2020

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

Tự tìm Đkxđ nha.

1/(3y^2 - 10y +3) = 6y/(9y^2 - 1) + 2/(1 - 3y)

=>1/(3y^2 -9y -y +3)=6y/(3y- 1)(3y+ 1)- 2(3y+ 1)/(3y - 1)(3y+ 1)

=>1/(y- 3)(3y -1)=-1/(3y -1)(3y +1)

=>(3y+ 1)/(y- 3)(3y -1)(3y+ 1)=(y -3)/(3y- 1)(3y +1)

=>3y+ 1= y- 3

Đến đây tự làm nha

21 tháng 2 2019

a)ĐKXĐ:\(\hept{\begin{cases}y\ne3\\y\ne\frac{1}{3}\\y\ne-\frac{1}{3}\end{cases}}\)

\(\frac{1}{3y^2-10y+3}=\frac{6y}{9y^2-1}+\frac{2}{1-3y}\)

\(\Leftrightarrow\frac{1}{\left(y-3\right)\left(3y-1\right)}=\frac{6y}{\left(3y-1\right)\left(3y+1\right)}-\frac{2}{3y-1}\)

\(\Leftrightarrow\frac{3y+1}{\left(y-3\right)\left(3y-1\right)\left(3y+1\right)}=\frac{6y\left(y-3\right)}{\left(3y-1\right)\left(3y+1\right)\left(y-3\right)}-\frac{2\left(3y+1\right)\left(y-3\right)}{\left(3y-1\right)\left(3y+1\right)\left(y-3\right)}\)

\(\Rightarrow6y^2-18y-2\left(3y^2-9y+y-3\right)-3y-1=0\)

\(\Leftrightarrow6y^2-18y-6y^2+18y-2y+6-3y-1=0\)

\(\Leftrightarrow5-5y=0\)

\(\Leftrightarrow5y=5\Leftrightarrow y=1\)(t/m ĐKXĐ)

Vậy....

19 tháng 7 2016

\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

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

 

19 tháng 7 2016

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

26 tháng 12 2020

a. 2x(x + y) - y(y + 2x) = 2x2 + 2xy - y2 - 2xy = 2x2 - y2

b.\(\frac{4x+3y}{7x^2y}-\frac{3x+3y}{7x^2y}=\frac{4x+3y-3x-3y}{7x^2y}=\frac{x}{7x^2y}=\frac{1}{7xy}\)

Phần c nản quá.