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 20/07/2021 Bởi Camila 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: $a,6x^{3}+x^2=2x$ ⇔$6x^{3}+x^2-2x=0$ ⇔$x(6x^{2}+x-2)=0$ ⇔\(\left[ \begin{array}{l}x=0\\6x^2+x-2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\(2x-1)(3x+2)=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\x=\frac{1}2\\x=\frac{-2}{3}\end{array} \right.\) $b,x^{2}+5x+4=0$ ⇔$x^{2}+4x+x+4=0$ ⇔$(x+1)(x+4)=0$ ⇔\(\left[ \begin{array}{l}x+1=0\\x+4=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\) $c,2x^{2}+3x-5=0$ ⇔$2x^{2}+5x-2x-5=0$ ⇔$(x-1)(2x+5)=0$ ⇔\(\left[ \begin{array}{l}x-1=0\\2x+5=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-5}2\end{array} \right.\) $d,x^{2}+5x=6$ ⇔$x^{2}+5x-6=0$ ⇔$x^{2}+6x-x-6=0$ ⇔$(x-1)(x+6)=0$ ⇔\(\left[ \begin{array}{l}x-1=0\\x+6=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=-6\end{array} \right.\) $e,16x-5x^{2}=3$ ⇔$-5x^{2}+16x-3=0$ ⇔$5x^{2}-16x+3=0$ ⇔$5x^{2}-x-15x+3=0$ ⇔$(x-3)(5x-1)=0$ ⇔\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=3\\x=\frac{1}5\end{array} \right.\) Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
Giải thích các bước giải:
$a,6x^{3}+x^2=2x$
⇔$6x^{3}+x^2-2x=0$
⇔$x(6x^{2}+x-2)=0$
⇔\(\left[ \begin{array}{l}x=0\\6x^2+x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\(2x-1)(3x+2)=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=\frac{1}2\\x=\frac{-2}{3}\end{array} \right.\)
$b,x^{2}+5x+4=0$
⇔$x^{2}+4x+x+4=0$
⇔$(x+1)(x+4)=0$
⇔\(\left[ \begin{array}{l}x+1=0\\x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=4\end{array} \right.\)
$c,2x^{2}+3x-5=0$
⇔$2x^{2}+5x-2x-5=0$
⇔$(x-1)(2x+5)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\2x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=\frac{-5}2\end{array} \right.\)
$d,x^{2}+5x=6$
⇔$x^{2}+5x-6=0$
⇔$x^{2}+6x-x-6=0$
⇔$(x-1)(x+6)=0$
⇔\(\left[ \begin{array}{l}x-1=0\\x+6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-6\end{array} \right.\)
$e,16x-5x^{2}=3$
⇔$-5x^{2}+16x-3=0$
⇔$5x^{2}-16x+3=0$
⇔$5x^{2}-x-15x+3=0$
⇔$(x-3)(5x-1)=0$
⇔\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=\frac{1}5\end{array} \right.\)