Toán x^3+x=0 2x(x-9)+3(x-9) =0 6x^2-3x=0 5x^3(7x+1)-10x^2(7x+1)=0 09/07/2021 By Ruby x^3+x=0 2x(x-9)+3(x-9) =0 6x^2-3x=0 5x^3(7x+1)-10x^2(7x+1)=0
Đáp án + Giải thích các bước giải: `a,x^3+x=0` `=> x(x^2+1)=0` \(⇒\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=0\quad(Tm)\\x^2=-1\quad(Ktm)\end{array} \right.\) Vậy `S={0}` `b,2x(x-9)+3(x-9)=0` `=> (x-9)(2x+3)=0` \(⇒\left[ \begin{array}{l}x-9=0\\2x+3=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=9\\2x=-3\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.\) Vậy `S={9;-3/2}` `c,6x^2-3x=0` `=>3x(2x-1)=0` \(⇒\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=0\\2x=1\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.\) Vậy `S={0;1/2}` `d,5x^3(7x+1)-10x^2(7x+1)=0` `=>(7x+1)(5x^3-10x^2)=0` `=>(7x+1)(x-2)5x^2=0` \(⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.\) Vậy `S={-1/7;2;0}` Trả lời
`x^3+x=0` `⇒x(x^2+1)=0` `⇒` \(\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\) `⇒` \(\left[ \begin{array}{l}x=0\\x^2=-1\text{ (loại; do x² ≥ 0)}\end{array} \right.\) Vậy `x=0`. `2x(x-9)+3(x-9) =0` `⇒(2x+3)(x-9)=0` `⇒` \(\left[ \begin{array}{l}2x+3=0\\x-9=0\end{array} \right.\) $⇒ \left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.$ Vậy `x=-3/2` hoặc `x=9`. `6x^2-3x=0` `⇒3x(2x-1)=0` `⇒` $\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.$ Vậy `x=0` hoặc `x=1/2`. `5x^3(7x+1)-10x^2(7x+1)=0` `⇒(7x+1)(5x^3-10x^2)=0` `⇒(7x+1)(x-2)5x^2=0` $⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.$ $⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.$ Vậy `x∈{-1/7;2;0}`. Trả lời
Đáp án + Giải thích các bước giải:
`a,x^3+x=0`
`=> x(x^2+1)=0`
\(⇒\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\quad(Tm)\\x^2=-1\quad(Ktm)\end{array} \right.\)
Vậy `S={0}`
`b,2x(x-9)+3(x-9)=0`
`=> (x-9)(2x+3)=0`
\(⇒\left[ \begin{array}{l}x-9=0\\2x+3=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=9\\2x=-3\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.\)
Vậy `S={9;-3/2}`
`c,6x^2-3x=0`
`=>3x(2x-1)=0`
\(⇒\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\2x=1\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.\)
Vậy `S={0;1/2}`
`d,5x^3(7x+1)-10x^2(7x+1)=0`
`=>(7x+1)(5x^3-10x^2)=0`
`=>(7x+1)(x-2)5x^2=0`
\(⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.\)
Vậy `S={-1/7;2;0}`
`x^3+x=0`
`⇒x(x^2+1)=0`
`⇒` \(\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x=0\\x^2=-1\text{ (loại; do x² ≥ 0)}\end{array} \right.\)
Vậy `x=0`.
`2x(x-9)+3(x-9) =0`
`⇒(2x+3)(x-9)=0`
`⇒` \(\left[ \begin{array}{l}2x+3=0\\x-9=0\end{array} \right.\)
$⇒ \left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.$
Vậy `x=-3/2` hoặc `x=9`.
`6x^2-3x=0`
`⇒3x(2x-1)=0`
`⇒` $\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.$
Vậy `x=0` hoặc `x=1/2`.
`5x^3(7x+1)-10x^2(7x+1)=0`
`⇒(7x+1)(5x^3-10x^2)=0`
`⇒(7x+1)(x-2)5x^2=0`
$⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.$
$⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.$
Vậy `x∈{-1/7;2;0}`.