Suy ra (2x-4)-(3x-3×5)=1 Suy ra(2x-4)-3x+15=1 Suy ra 2x-4-3x+15=1 Suy ra (2x-3x)+(15-4)=1 -1x+11=1 1-11=-1x -1x=-10 X=10
x-3)^2-2(2x-7)(x-3)=0
Answer:
1, \(x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow x\left(2x-7\right)-2\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3,5\\x=2\end{cases}}}\)
2, \(x+x^2-x^3-x^4=0\)
\(\Leftrightarrow x\left(1+x-x^2-x^3\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)^2.\left(x-1\right)=0\)
Trường hợp 1: \(x=0\)
Trường hợp 2: \(x+1=0\Leftrightarrow x=-1\)
Trường hợp 3: \(x-1=0\Leftrightarrow x=1\)
3, \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+3x^2\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow x^2.\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
Vì \(x^2\ge0\forall x\)
\(\Leftrightarrow x^2+1\ge1\forall x\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
4, \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow\left(-2\right)\left(2x-5\right)=0\)
\(\Leftrightarrow2x=5\)
\(\Leftrightarrow x=\frac{2}{5}\)