Giải phương trình:
\(\frac{2x-1}{2}+\frac{5-x}{6}=2-\frac{3\left(x-1\right)}{4}\)
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\(\frac{x}{3} + \frac{{2x + 1}}{6} = \frac{{4\left( {x - 2} \right)}}{5}\)
\(\frac{{10x}}{{3.10}} + \frac{{\left( {2x + 1} \right).5}}{{6.5}} = \frac{{6.4\left( {x - 2} \right)}}{{5.6}}\)
\(\frac{{10x}}{{30}} + \frac{{10x + 5}}{{30}} = \frac{{24x - 48}}{{30}}\)
\(10x + 10x + 5 = 24x - 48\)
\(10x + 10x - 24x = - 5 - 48\)
\( - 4x = - 53\)
\(x = \left( { - 53} \right):\left( { - 4} \right)\)
\(x = \frac{{53}}{4}\)
Vậy phương trình có nghiệm là \(x = \frac{{53}}{4}\).
a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)
\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)
\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)
\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
còn đây là câu b
\(\frac{3x-2-30}{6}=\frac{3-2x-14}{4}\)
\(\Leftrightarrow\frac{3x-32}{6}-\frac{-11-2x}{4}=0\)
\(\Leftrightarrow\frac{6x-64+33+6x}{12}\)
\(\Leftrightarrow12x=31\)
\(\Leftrightarrow x=\frac{31}{12}\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)