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6 tháng 11 2016

mk ko biết làm 

xin lỗi bn nhae

xin lỗi vì đã ko giúp được bn

chcus bn học gioi!

nhae@@@

6 tháng 11 2016

mình không biết làm

tk nhé@@@@@@@@@@@@@@@@@@@@

LOL

hihi

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

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

\(x\left(x+1\right)+x\left(x-3\right)=4x\)

\(x^2+x+x^2-3x=4x\)

\(2x^2-2x=4x\)

\(2x^2-2x-4x=0\)

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

\(2x=0\Leftrightarrow x=0\)

hoặc 

\(x-3=0\Leftrightarrow x=3\)

22 tháng 4 2020

b) \(ĐKXĐ:x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )

Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)

11 tháng 12 2016

a) \(\frac{x-1}{x+1}-\frac{x+1}{x-1}+\frac{4}{x^2-1}\left(ĐK:x\ne\pm1\right)\)

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

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

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

b) \(\frac{x^3y+xy^3}{x^4y}:\left(x^2+y^2\right)\left(ĐK:x,y\ne0\right)\)

\(=\frac{xy\left(x^2+y^2\right)}{x^4y}\cdot\frac{1}{x^2+y^2}\)

\(=\frac{1}{x^3}\)

Bài 1: 

a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

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

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

b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)

c: Đề thiếu rồi bạn

9 tháng 12 2019

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

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

\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)

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

\(=\frac{3x-6}{x+2}\)

9 tháng 12 2019

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

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

\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)

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

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

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

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