\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(2x^2+3\left(x^2-1^2\right)=5x^2+5x\)
\(2x^2+3x^2-3=5x^2+5x\)
\(-3=5x\)
\(x=\frac{-5}{3}\)
2x2 + 3(x - 1)(x + 1) = 5x(x + 1)
=> 2x2 + 3(x2 - 1) = 5x2 + 5x
=> 2x2 + 3x2 - 3 = 5x2 + 5x
=> 5x2 - 3 = 5x2 + 5x
=> 5x = -3
=> x = -0,6
Vậy x = -0,6
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(2x^2+3x^2-3=5x^2+5x\)
\(-3=5x\)
\(x=\frac{-3}{5}\)
2x2 + 3(x - 1)(x + 1) = 5x(x + 1)
Áp dụng (A + B)(A - B) = A2 - B2 ( hằng đẳng thức số 3)
=> 2x2 + 3(x2 - 12) = 5x2 + 5x
=> 2x2 + 3x2 - 3 = 5x2 + 5x
=> 5x2 - 3 - 5x2 - 5x = 0
=> -3 - 5x = 0
=> 5x = -3
=> x = -3/5
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+3\left(x^2-1\right)-\left(5x^2+5x\right)=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)\(\Leftrightarrow5x=-3\)\(\Leftrightarrow x=\frac{-3}{5}\)
Vậy \(x=\frac{-3}{5}\)
2x^2+3(x-1)(x+1)=5x(x+1)
<=>2x 2 +3.(x2-1)=5x2-5x
<=>2x2+3x2-3-5x2+5x=0
<=>5x-3=0
<=>x=3/5
vậy tập nghiệm của pt là :S={3/5}