Question

Giải các phương trình:

\(a,4\left(x+3\right)^2=\left(2x+6\right)^2\)

\(b,x^3-3x+2=0\)

\(c,2x^3+5x^2=7x\)

Answer 1

a) 4(x+3)²=(2x+6)²

<=> 4(x²+6x+9)=4x²+24x+36

<=>4x²+24x+36=4x²+24+36

<=> 0x=0

=> x∈R

Vậy pt có nghiệm là : x ∈ R

Answer 2

\(x^3-3x+2=0\)

\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(2x^3+5x^2=7x\)

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

\(\Leftrightarrow x\left[2x\left(x-1\right)+7\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x+7\right)=0\)

Tự làm nốt