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17 tháng 12 2017

\(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)

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

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

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

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

18 tháng 4 2017

1.

\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\left(ĐKXĐ:x\ne1\right)\\ \Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\\ \Leftrightarrow21x-9=2x-2\\ \Leftrightarrow19x=7\\ \Leftrightarrow x=\dfrac{7}{19}\left(TMĐK\right)\)

2.

\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\left(ĐKXĐ:x\ne-\dfrac{2}{3};x\ne\dfrac{1}{3}\right)\\ \Leftrightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\\ \Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\\ \Leftrightarrow-8x+1=-11x-14\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\left(TMĐK\right)\)

3.

\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ \Leftrightarrow\left(\dfrac{1-x}{x+1}+3\right)\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{1-x+3\left(x+1\right)}{x+1}.\left(x+1\right)=2x+3\\ \Leftrightarrow\dfrac{4+2x}{x+1}\left(x+1\right)=2x+3\\ \Leftrightarrow4+2x=2x+3\\ \Leftrightarrow4=3\)

Vô nghiệm.

8 tháng 12 2017

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

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

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

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

16 tháng 11 2022

sai

 

 

 

23 tháng 2 2019

Câu 1:

Hỏi đáp Toán

23 tháng 2 2019

Câu 2:

ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

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

\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)

\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)

\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)

\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)

Vậy \(S=\left\{-1\right\}\)

13 tháng 11 2017

a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)+z\left(x-y\right)}\)

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

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

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

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

13 tháng 11 2017

a) áp dụng hằng đẳng thức

22 tháng 7 2017

a) \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)

\(=-\dfrac{9x^2+3x+2x-6x^2}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=-\dfrac{x\left(3x+5\right)}{\left(3x-1\right)^2}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

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

b) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

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

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

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

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

28 tháng 2 2018

Hỏi đáp ToánHỏi đáp Toán

25 tháng 2 2018

\(\dfrac{2x-8}{6}-\dfrac{3x+1}{4}=\dfrac{9x-2}{8}+\dfrac{3x-1}{12}\)

\(\Leftrightarrow4\left(2x-8\right)-6\left(3x+1\right)=3\left(9x-2\right)+2\left(3x-1\right)\)

\(\Leftrightarrow8x-32-18x-6=27x-6+6x-2\)

\(\Leftrightarrow8x-18x-27x-6x=-6-2+32+6\)

\(\Leftrightarrow-43x=30\)

\(\Leftrightarrow x=\dfrac{-30}{43}\)

\(\Rightarrow S=\left\{\dfrac{-30}{43}\right\}\)

11 tháng 11 2017

Nguyễn Ngọc Thanh Trúc đề là gì

11 tháng 11 2017

thực hiện phép tính