K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

16 tháng 10 2016

\(\frac{9x^2-12x^2+3x}{-3x}+3x\left(x-2\right)=\frac{-3x^2+3x}{-3x}+3x\left(x-2\right)=\frac{-3x\left(x-1\right)}{-3x}+3x^2-6x=x-1+3x^2-6x\)

\(=3x^2-5x-1\)

5 tháng 1 2022

\(a,\dfrac{3x+21}{x^2-9}+\dfrac{2}{x+3}-\dfrac{3}{x-3}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+21+2x-6-3x-9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2}{x-3}\)

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

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

9 tháng 11 2021

\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)-\left(x-1\right)}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\left(xy\ne1;x\ne1\right)\)

3 tháng 6 2021

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

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

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

\(\dfrac{3x^3+8x^2+7x+2}{\left(x^2-1\right)^2}-\dfrac{6x^2-6}{\left(x^2-1\right)^2}-\dfrac{3x^3-8x^2+7x-2}{\left(x^2-1\right)^2}\)

\(\dfrac{10x^2+10}{\left(x^2-1\right)^2}\)

\(\dfrac{10\left(x^2+1\right)}{\left(x^2-1\right)^2}\)

8 tháng 12 2021

ĐK: \(3x\ne\pm y;x\ne0\)

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

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

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)

10 tháng 10 2016

\(A=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)

    \(=\left[\left(3x^3+1\right)+3x\right]\left[\left(3x^3+1\right)-3x\right]-\left(3x^3+1\right)^2\)

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

    \(=-\left(3x\right)^2\)

     \(=-9x^2\)

10 tháng 10 2016

\(\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2=\left(3x^3+1\right)^2-\left(3x\right)^2-\left(3x^3+1\right)^2=-9x^2\)

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

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x-3}-\dfrac{6x+9}{2x+3}\cdot\dfrac{1}{x-3}\)

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

 

9 tháng 12 2021

\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)

4 tháng 10 2016

\(A=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)

\(=\left[\left(3x^3+1\right)+3x\right]\left[\left(3x^3+1\right)-3x\right]-\left(3x^3+1\right)^2\)

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

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

\(=-9x^2\)

16 tháng 7 2016

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

=>\(C=\frac{x}{x-3}-\frac{x\left(x+3\right)}{2x+3}.\left[\frac{x+3}{x\left(x-3\right)}-\frac{x}{\left(x-3\right)\left(x+3\right)}\right]\)

=>\(C=\frac{x}{x-3}-\frac{x\left(x+3\right)}{2x+3}\left[\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}-\frac{x^2}{x\left(x-3\right)\left(x+3\right)}\right]\)

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

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

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

=>\(C=\frac{x}{x-3}-\frac{3}{x-3}\)

=>\(C=\frac{x-3}{x-3}\)

=>C=1