Question

Giải các phương trình sau:

\(\left(x-2\right)\left(x+1\right)=x^2-4\).

\(x^2-5x+6=0\).

\(2x^{^3}+6x^2=x^2+3x\). Giúp mik nha, mik đang cần rất gấp

Answer 1

\(x^2-5x+6=\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)

Answer 2

1, <=>x^2-x-2 = x^2-4

<=>x^2-4-x^2+x+2 = 0

<=> x-2 = 0

<=> x=2

2, <=> (x-2).(x-3)=0

<=> x-2 = 0 hoặc x-3 = 0

<=> x=2 hoặc x=3

Answer 3

\(\left(x-2\right)\left(x+1\right)=x^2-x-2=x^2-4\)

\(\Leftrightarrow\left(-x\right)-2=-4\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\)

Answer 4

Ta có

\(a,\left(x-2\right)\left(x+1\right)=x^2-4\)

\(\left(x-2\right)\left(x+1\right)-x^2+4=0\)

\(x^2-2x+x-2-x^2+4=0\)

\(-x+2=0\Leftrightarrow x=2\)

\(b,x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)

\(\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(c,2x^3+6x^2=x^2+3x\Leftrightarrow2x^3+6x^2-x^2-3x=0\Leftrightarrow2x^3+5x^2-3x=0\)

\(x\cdot\left(2x-1\right)\cdot\left(x+3\right)=0\Rightarrow x=0;x=\frac{1}{2};x=-3\)