số gt x thỏa mãn\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
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Ta có;
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}=\frac{3x-1+25-3x}{40-5x+5x-34}=\frac{24}{6}=4\)
=> 25 - 3x = 4(5x - 34)
=> 25 - 3x = 20x - 136
=> 25 + 136 = 20x + 3x
=> 161 = 23x
=> x = 7
với đề là (3x - 1)/40 - 5x = (25 -3x)/5x-3 thì nhân chéo lên mà giải thui mà:
(3x -1).(5x -34 )=(40 - 5x).(25 - 3x)
=>x=7
Ta có :
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(=>\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(=>15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(=>15x^2-107x+34=1000-245x+15x^2\)
\(=>34-107x=1000-245x\)
\(=>1000-245x+107x=34\)
\(=>1000-138x=34\)
\(=>138x=1000-34=966\)
\(=>x=\frac{966}{138}=7\)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(\Rightarrow3x\left(5x-34\right)-\left(5x-34\right)=40\left(25-3x\right)-5x\left(25-3x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Rightarrow15x^2-97x+34=1000-245x+15x^2\)
\(\Rightarrow15x^2=1000-34-245x+97x+15^2\)
\(\Rightarrow15x^2=966-148x+15^x\)
\(\Rightarrow0=966-148x\)
\(\Rightarrow x=\frac{996}{148}=\frac{249}{37}\)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
\(x\ne8;x\ne\frac{34}{5}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\)
\(\Rightarrow-102x-5x+125x+120x=1000-34\)
\(\Rightarrow138x=966\Rightarrow x=7\)
Vậy x = 7
x = 7
tick mình nha!