3x-1/40-5x=25-3x/5x-34
3x-1/40-5x=25-3x/5x-34
(3x-1)/(40-5x)=(25-3x)/(5x-34)
Tìm x
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)
\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)
\(\Leftrightarrow138x=966\)
\(\Leftrightarrow x=966:138\)
\(\Leftrightarrow x=7\)
Tìm x :
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
Ta có :
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(=>\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(=>15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(=>15x^2-107x+34=1000-245x+15x^2\)
\(=>34-107x=1000-245x\)
\(=>1000-245x+107x=34\)
\(=>1000-138x=34\)
\(=>138x=1000-34=966\)
\(=>x=\frac{966}{138}=7\)
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=\(\pm\sqrt{7}\)
Tìm x biết:
a) (2x + 3)/(5x +2) = (4x + 5)/(10x + 2)
b) (3x - 1)/(40 - 5x) = (25 - 3x)/(5x - 34)
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
Tìm x:
\(\dfrac{3x-1}{40-5x}=\dfrac{25-3x}{5x-34}\)
\(\dfrac{3x-1}{40-5x}=\dfrac{25-3x}{5x-34}\\ \Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\\ \Leftrightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\\ \Leftrightarrow15x^2-15x^2-102x-5x+125x+120x=1000-34\\ \Leftrightarrow138x=966\\ \Leftrightarrow x=\dfrac{966}{138}\\ \Leftrightarrow x=7\)
Tìm x biết :
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)
\(\Rightarrow3x\left(5x-34\right)-\left(5x-34\right)=40\left(25-3x\right)-5x\left(25-3x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)
\(\Rightarrow15x^2-97x+34=1000-245x+15x^2\)
\(\Rightarrow15x^2=1000-34-245x+97x+15^2\)
\(\Rightarrow15x^2=966-148x+15^x\)
\(\Rightarrow0=966-148x\)
\(\Rightarrow x=\frac{996}{148}=\frac{249}{37}\)
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\) giải hộ mình bài này nha
áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}=\frac{3x-1+25-3x}{40-5x+5x-34}=\frac{24}{6}=4\)
suy ra:
\(\frac{3x-1}{40x-5}=4\Rightarrow3x-1=4.\left(40-5x\right)\)
3x-1=160-20x
3x+20x=160+1
23x=161
x=161:23
x=7
vậy x=7
số gt x thỏa mãn\(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)
\(x\ne8;x\ne\frac{34}{5}\)
\(\Rightarrow\left(3x-1\right)\left(5x-34\right)=\left(25-3x\right)\left(40-5x\right)\)
\(\Rightarrow15x^2-102x-5x+34=1000-125x-120x+15x^2\)
\(\Rightarrow-102x-5x+125x+120x=1000-34\)
\(\Rightarrow138x=966\Rightarrow x=7\)
Vậy x = 7