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26 tháng 10 2023

a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)

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

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

b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)

\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(8-x\right)}{x+2}\)

\(=\dfrac{x-8}{x+2}\)

c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)

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

\(=-2\)

d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

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

26 tháng 10 2023

ĐKXĐ: \(x\neq0;x\neq-1\)

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

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)

$---$

ĐKXĐ: \(x\neq y\)

\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)

$---$

ĐKXĐ: \(x\neq-2\)

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

28 tháng 6 2017

Quy đồng mẫu thức nhiều phân thức

Quy đồng mẫu thức nhiều phân thức

17 tháng 11 2017

Bạn siêng thật !!!

a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)

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

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

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

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

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

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

10 tháng 12 2018

1.

a) \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

b) \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

10 tháng 12 2018

Bài 1:

a, \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

Vậy \(x=-4\) hoặc \(x=-1\)

b, \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x=3\) hoặc \(x=-2\)

28 tháng 6 2017

Phép nhân các phân thức đại số

1, Thực hiện phép tính : a, \(\dfrac{2x+4}{10}\) + \(\dfrac{2-x}{15}\) b, \(\dfrac{3x}{10}\) + \(\dfrac{2x-1}{15}\) + \(\dfrac{2-x}{20}\) c, \(\dfrac{x+1}{2x-2}\) + \(\dfrac{x^2+3}{2-2x^2}\) d, \(\dfrac{1-2x}{2x}\) + \(\dfrac{2x}{2x-1}\) + \(\dfrac{1}{2x-4x^2}\) e, \(\dfrac{x}{xy-y^2}\) + \(\dfrac{2x-y}{xy-x^2}\) f, \(\dfrac{x^2}{x^2-4x}\) + \(\dfrac{6}{6-3x}\) +\(\dfrac{1}{x+2}\) g, \(\dfrac{2x^2-10xy}{2xy}\) + \(\dfrac{5y-x}{y}\) + \(\dfrac{x+2y}{x}\) h, \(\dfrac{2}{x+y}\)...
Đọc tiếp

1, Thực hiện phép tính :

a, \(\dfrac{2x+4}{10}\) + \(\dfrac{2-x}{15}\)

b, \(\dfrac{3x}{10}\) + \(\dfrac{2x-1}{15}\) + \(\dfrac{2-x}{20}\)

c, \(\dfrac{x+1}{2x-2}\) + \(\dfrac{x^2+3}{2-2x^2}\)

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

e, \(\dfrac{x}{xy-y^2}\) + \(\dfrac{2x-y}{xy-x^2}\)

f, \(\dfrac{x^2}{x^2-4x}\) + \(\dfrac{6}{6-3x}\) +\(\dfrac{1}{x+2}\)

g, \(\dfrac{2x^2-10xy}{2xy}\) + \(\dfrac{5y-x}{y}\) + \(\dfrac{x+2y}{x}\)

h, \(\dfrac{2}{x+y}\) +\(\dfrac{1}{x-y}\) + \(\dfrac{-3x}{x^2-y^2}\)

i, x+y+ \(\dfrac{x^2+y^2}{x+y}\)

2, Thực hiện phép tính :

a, \(\dfrac{2x}{x^2+2xy}\) + \(\dfrac{y}{xy-2y^2}\)+ \(\dfrac{4}{x^2-4y^2}\)

b, \(\dfrac{1}{x-y}\) + \(\dfrac{3xy}{y^3-x^3}\) + \(\dfrac{x-y}{x^2+xy+y^2}\)

c, \(\dfrac{2x+y}{2x^2-xy}\) + \(\dfrac{16x}{y^2-4x^2}\) + \(\dfrac{2x-y}{2x^2+xy}\)

d, \(\dfrac{1}{1-x}\) +\(\dfrac{1}{1+x}\) + \(\dfrac{2}{1+x^2}\) + \(\dfrac{4}{1+x^4}\) + \(\dfrac{8}{1+x^8}\)+ \(\dfrac{16}{1+x^{16}}\)

1
13 tháng 11 2017

Bài 2 .

a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)

\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)

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

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

\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)

b) Sai đề hay sao ý

c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)

\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)

\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)

\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)

d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

.....

\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{32}{1-x^{32}}\)

19 tháng 12 2018

a) \(\dfrac{x^2-4}{2x^2-4x}\)

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

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

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

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

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

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

c) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

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

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

\(=\dfrac{x-y}{x+y}\)

d) \(\dfrac{5x^2+10x+5}{x+x^2}\)

\(\dfrac{5\left(x^2+2x+1\right)}{x\left(1+x\right)}\)

\(=\dfrac{5\left(x+1\right)^2}{x\left(x+1\right)}\)

\(=\dfrac{5\left(x+1\right)}{x}\)

e) \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x+6\right)}\)

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

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

f) \(\dfrac{\left(2-3x\right)\left(x+1\right)}{x^2+2x+1}\)

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

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

1 tháng 12 2017

a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

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

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

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

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

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

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

\(=\dfrac{1}{x}\)

b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)

\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)

\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)

\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)

1 tháng 12 2017

c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)

\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)

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

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

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

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

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

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

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

d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)

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

\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)

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

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

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

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

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

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

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

b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)

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

\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)

 

20 tháng 11 2018

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

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

20 tháng 11 2018

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

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

15 tháng 11 2017

2)

a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)

\(=\dfrac{6x}{xy}\)

\(=\dfrac{6}{y}\)

b) \(\dfrac{2x^2}{y}.3xy^2\)

\(=\dfrac{2x^2.3xy^2}{y}\)

\(=\dfrac{6x^3y^2}{y}\)

\(=6x^3y\)

c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)

\(=\dfrac{15x.2y^2}{7y^3.x^2}\)

\(=\dfrac{30xy^2}{7x^2y^3}\)

\(=\dfrac{30}{7xy}\)

d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)

\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)

\(=\dfrac{2y}{5x\left(x-y\right)}\)