Question

Giải các phương trình sau :

1.(2x+1).(x²+2)=0

2.(x²+4).(7x-3)=0

3.(x-5).(3-2x).(3x+4)=0

4.(x-2).(3x+5)=(2x-4).(x+1)

5.(2x+5).(x-4)=(x-5).(4-x)

Giúp mình với

Answer 1

1.

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

\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{2}\\ x^2=-2< 0(\text{vô lý})\end{matrix}\right.\)

Vậy \(x=-\frac{1}{2}\)

2.\((x^2+4)(7x-3)=0\Rightarrow \left[\begin{matrix} x^2+4=0\\ 7x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2=-4< 0(\text{vô lý})\\ x=\frac{3}{7}\end{matrix}\right.\)

Vậy \(x=\frac{3}{7}\)

3.

\((x-5)(3-2x)(3x+4)=0\)

\(\Rightarrow \left[\begin{matrix} x-5=0\\ 3-2x=0\\ 3x+4=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=\frac{3}{2}\\ x=-\frac{4}{3}\end{matrix}\right.\)

Answer 2

4.

\((x-2)(3x+5)=(2x-4)(x+1)\)

\(\Leftrightarrow (x-2)(3x+5)-(2x-4)(x+1)=0\)

\(\Leftrightarrow (x-2)(3x+5)-2(x-2)(x+1)=0\)

\(\Leftrightarrow (x-2)[(3x+5)-2(x+1)]=0\)

\(\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x-2=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

5.

\((2x+5)(x-4)=(x-5)(4-x)\)

\(\Leftrightarrow (2x+5)(x-4)-(x-5)(4-x)=0\)

\(\Leftrightarrow (2x+5)(x-4)+(x-5)(x-4)=0\)

\(\Leftrightarrow (x-4)[(2x+5)+(x-5)]=0\)

\(\Leftrightarrow (x-4).3x=0\)

\(\Rightarrow \left[\begin{matrix} x-4=0\\ 3x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=4\\ x=0\end{matrix}\right.\)