Question

 1, x(2x-7)-4x+14=0           2, x+x^2-x^3-x^4=0

 3, 2x^3+3x^2+2x+3=0      4, 4x^2 - 25 - (2x-5)(2x+7)=0

 

Answer 1

mk ko bt 123

Answer 2

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

Answer 3

x-3)^2-2(2x-7)(x-3)=0

Answer 4

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, \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow\left(-2\right)\left(2x-5\right)=0\)

\(\Leftrightarrow2x=5\)

\(\Leftrightarrow x=\frac{2}{5}\)