1) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
2)\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)
3) \(\frac{5}{2x-4}+\frac{7}{x+2}+\frac{-1}{x^2-4}\)
4) \(\frac{x+3}{x^2+x-2}+\frac{4-x}{x^2+5x+6}\)
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Những câu hỏi liên quan
27 tháng 4 2020
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn
NV
18 tháng 4 2017
\(1.\frac{7x-3}{x-1}=\frac{2}{3}\) ( \(x\ne1\))
\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)
\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)
\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-5\)
\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)
\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)
\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)
\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)
\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)
\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)
\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)
\(\Leftrightarrow4x^2+5x-7=0\)
\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)
\(\left(2x+\frac{5}{4}\right)^2>0\)
\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)
=> PT vô nghiệm
\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\)
\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)
\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{16}{6}\)
\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)
\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)
\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)
\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)
\(\Leftrightarrow x^4+x^3-4x-8=0\)
\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)
Đến đấy mk tắc r xl bạn nhé
22 tháng 4 2020
Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
23 tháng 4 2020
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
24 tháng 6 2019
a,ĐKXĐ \(x\ne-1;-\frac{1}{2}\)
Ta thấy x=0 không là nghiệm của PT
Xét \(x\ne0\)
Khi đó PT
<=> \(\frac{2}{6x-1+\frac{3}{x}}+\frac{5}{4x+5+\frac{2}{x}}+\frac{1}{2x+3+\frac{1}{x}}=\frac{1}{3}\)
Đặt \(2x+\frac{1}{x}=a\)
=> \(\frac{2}{3a-1}+\frac{5}{2a+5}+\frac{1}{a+3}=\frac{1}{3}\)
<=> \(3\left(25a^2+75a+10\right)=6a^3+31a^2+34a-15\)
<=> \(6a^3-44a^2-191a-45=0\)
Xin lỗi đến đây tớ ra nghiệm không đẹp
24 tháng 6 2019
c, \(x^2+\frac{9x^2}{\left(x+3\right)^2}=7\) ĐKXĐ \(x\ne-3\)
<=> \(\left(x-\frac{3x}{x+3}\right)^2+2.\frac{3x^2}{x+3}=7\)
<=> \(\left(\frac{x^2}{x+3}\right)^2+6.\frac{x^2}{x+3}-7=0\)
<=> \(\left(\frac{x^2}{x+3}+7\right)\left(\frac{x^2}{x+3}-1\right)=0\)
<=> \(\orbr{\begin{cases}x^2+7x+21=0\\x^2-x-3=0\end{cases}}\)
\(S=\left\{\frac{1\pm\sqrt{13}}{2}\right\}\)thỏa mãn ĐKXĐ
1)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3x}{\left(2x+6\right)x}-\frac{x-6}{2x^2+6x}\\ =\frac{3x}{2x^2+6x}-\frac{x-6}{2x^2+6x}=\frac{3x-\left(x-6\right)}{2x^2+6x}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)