Ta có :
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(=>\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(=>15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(=>15x^2-107x+34=1000-245x+15x^2\)
\(=>34-107x=1000-245x\)
\(=>1000-245x+107x=34\)
\(=>1000-138x=34\)
\(=>138x=1000-34=966\)
\(=>x=\frac{966}{138}=7\)
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theo đề bài ta có: \(\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
=> \(15x^2-107x+34=1000-245x+15x^2\)
<=> 138*x^2=966
<=>x^2=7
<=>x=\(\pm\sqrt{7}\)
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