Tìm x
(2x-1)2+(x+3)2-5.(x+7).(x-7)=0
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2x-\(\frac{1}{3}\)=1-\(\frac{5}{6}\)
2x-\(\frac{1}{3}\)=\(\frac{1}{6}\)
2x=\(\frac{1}{6}\)+\(\frac{1}{3}\)
2x=1/6 +2/6
2x=\(\frac{1}{2}\)
x=1/2 : 2
x/\(\frac{1}{4}\)
\(\frac{7}{9}\):(2+\(\frac{3}{4}\)x)+\(\frac{5}{9}\)=\(\frac{23}{27}\)
7/9 :(2+3/4x)=\(\frac{23}{27}\)-\(\frac{5}{9}\)
7/9 :(2+3/4x)=\(\frac{23}{27}\)-\(\frac{15}{27}\)
7/9 :(2+3/4x)=\(\frac{8}{27}\)
(2+3/4x) =\(\frac{7}{9}\) . \(\frac{27}{8}\)
(2+3/4x) =\(\frac{21}{8}\)
\(\frac{3}{4}\)x =\(\frac{21}{8}\)-2
3/4x =21/8 -16/8
3/4x = 5/8
x =\(\frac{5}{8}\) : \(\frac{3}{4}\)
x =5/8 . 4/3
x =\(\frac{20}{24}\)
= \(\frac{5}{2}-x=\frac{3}{5}+2x\)
=> x = \(\frac{19}{30}\)
x = \(0,6\left(3\right)\)
\(\frac{5}{2}-x=\frac{3}{5}+2x\)
\(-x-2x=\frac{3}{5}-\frac{5}{2}\)
\(-3x=\frac{-19}{10}\)
\(x=\frac{-19}{10}:\left(-3\right)\)
\(x=\frac{19}{30}\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\) ĐKXĐ: \(x\ne1;x\ne-7\)
\(\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-2x+7x-14=x^2-x+4x-4\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow x^2-x^2+5x-3x=-4+14\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)( thỏa mãn điều kiện xác định)
vậy x=5
Ta có:\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
\(\Rightarrow\left(x-2\right)\left(x-7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Rightarrow x^2-9x+14=x^2+3x-4\)
\(\Rightarrow x^2-9x+14-x^2-3x+4=0\)
\(\Rightarrow18-12x=0\)
\(\Rightarrow x=\frac{18}{12}\)