\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
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\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
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.
@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 \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
\(=\frac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}.\frac{5\left(x-1\right)}{3\left(x+1\right)}=\frac{x\left(x+1\right).5\left(x-1\right)}{5\left(x-1\right)^2.3\left(x+1\right)}=\frac{x}{3x-3}\)
\(\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\)
=\(\frac{x\left(x+1\right)}{5\left(x^2-2+1\right)}:\frac{3\left(x+1\right)}{5\left(x-1\right)}\)
=\(\frac{x\left(x+1\right)}{5\left(x-1\right)^2}:\frac{3\left(x+1\right)}{5\left(x-1\right)}\)
=\(\frac{x\left(x+1\right)}{5\left(x-1\right)^2}\cdot\frac{5\left(x-1\right)}{3\left(x+1\right)}\)
=\(\frac{x}{3\left(x-1\right)}\)
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
<=> \(\frac{\left(5x-2\right).2}{3.2}=\frac{\left(5-3x\right).3}{2.3}\)
<=> \(\frac{10x-4}{6}=\frac{15-9x}{6}\)
Suy ra: 10x - 4 = 15 - 9x
<=> 10x + 9x = 15 + 4
<=> 19x = 19
<=> x = 19:19
<=> x = 1
Vậy: S={1}
Trả lời :
\(\frac{5x-2}{3}=\frac{5-3x}{2}\)
\(\Rightarrow2\times\left(5x-2\right)=3\times\left(5-3x\right)\)
\(\Rightarrow10x-4=15-9x\)
\(\Rightarrow10x+9x=15+4\)
\(\Rightarrow19x=19\)
\(\Rightarrow x=1\)