Hộ mik luôn nhé Tìm x biết: a, 6x ³+x ²=2x b, x ²+ 5x+4=0 c, 2x ²+3x-5=0 d, x ²+5x=6 e, 16x-5x ²=3

Đáp án:

Giải thích các bước giải:

$a,6x^{3}+x^2=2x$ 

⇔$6x^{3}+x^2-2x=0$ 

⇔$x(6x^{2}+x-2)=0$ 

⇔\(\left[ \begin{array}{l}x=0\\6x^2+x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\(2x-1)(3x+2)=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=\frac{1}2\\x=\frac{-2}{3}\end{array} \right.\) 

$b,x^{2}+5x+4=0$

⇔$x^{2}+4x+x+4=0$ 

⇔$(x+1)(x+4)=0$

⇔\(\left[ \begin{array}{l}x+1=0\\x+4=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\) 

$c,2x^{2}+3x-5=0$

⇔$2x^{2}+5x-2x-5=0$ 

⇔$(x-1)(2x+5)=0$

⇔\(\left[ \begin{array}{l}x-1=0\\2x+5=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-5}2\end{array} \right.\) 

$d,x^{2}+5x=6$

⇔$x^{2}+5x-6=0$ 

⇔$x^{2}+6x-x-6=0$ 

⇔$(x-1)(x+6)=0$

⇔\(\left[ \begin{array}{l}x-1=0\\x+6=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=1\\x=-6\end{array} \right.\) 

$e,16x-5x^{2}=3$

⇔$-5x^{2}+16x-3=0$ 

⇔$5x^{2}-16x+3=0$ 

⇔$5x^{2}-x-15x+3=0$ 

⇔$(x-3)(5x-1)=0$

⇔\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=3\\x=\frac{1}5\end{array} \right.\)