Question

Giải các phương trình sau bằng cách đưa về phương trình tích

a) \({x} ^ {3}\)+ \({ x} ^ {2}\)- 12x =0

b) \({x} ^ {3} - { 4x} ^ {2} -x + 4 = 0\)

c) \(( { 2x} ^{2} -1) (3x-2) = ({ 2x} ^2 -1) ( x+ 3)\)

d) \(4x^2 - 12x +5 = 0\)

e) \(2x^2 + 5x + 3 = 0\)

Answer 1

a, x³ +x² -12x=0

\(\Leftrightarrow\)x³ +4x²-3x²-12x=0

\(\Leftrightarrow\) x²(x+4)-3x(x+4)=0

\(\Leftrightarrow\) (x²-3x)(x+4)=0

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

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

Vậy S\(=\)\(\left\{0;3;-4\right\}\)

Answer 2

b.x³-4x²-x+4=0

\(\Leftrightarrow\)x²(x-4)-(x-4)=0

\(\Leftrightarrow\) (x² -1)(x-4)=0

\(\Leftrightarrow\)(x-1)(x+1)(x-4)=0

\(\left[\begin{matrix}x+1=0\\x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=1\\x=-1\\x=4\end{matrix}\right.\)

Vậy S=\(\left\{1;-1;4\right\}\)

Answer 3

c. (2x²-1)(3x-2)=(2x²-1)(x+3)

\(\Leftrightarrow\)(2x²-1)(3x-2)-(2x²-1)(x+3)=0

\(\Leftrightarrow\)(2x²-1)(3x-2-x-3)=0

\(\Leftrightarrow\)(2x²-1)(2x-5)=0

\(\Leftrightarrow\)\(\left[\begin{matrix}2x^2-1=0\\2x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}2x^2=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x^2=\frac{1}{2}\\2x=5\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=\pm\sqrt{\frac{1}{2}}\\x=\frac{5}{2}\end{matrix}\right.\)

Answer 4

e, 2x²+5x+3=0

\(\Leftrightarrow\)2x² +2x+3x+3=0

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

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

\(\Leftrightarrow\) \(\left[\begin{matrix}2x+3=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}2x=-3\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{-3}{2}\\x=-1\end{matrix}\right.\)

Answer 5

cái này mình chưa được học nên không biết làm