4,5x^2 – 72x + 162 = 0 biết kết quả cuối cùng là x=18 23/10/2021 Bởi Everleigh 4,5x^2 – 72x + 162 = 0 biết kết quả cuối cùng là x=18
`4,5x^2 – 72x + 162 = 0` `<=>x^2 – 16x +36 = 0` `<=>x^2+2x – 18x +36 = 0` `<=>(x^2+2x) -( 18x +36) = 0` `<=>x(x+2) – 18(x +2) = 0` `<=>(x – 18)(x +2) = 0` \(⇔\left[ \begin{array}{l}x-18=0\\x+2=0\end{array} \right.\)\(⇔\left[ \begin{array}{l}x=18\\x=-2\end{array} \right.\) `tox={18;-2}` Bình luận
Cách 1: $\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5$ $⇔ \dfrac{36(x-6)+36(x+6)}{(x-6)(x+6)}=\dfrac{4,5(x-6)(x+6)}{(x-6)(x+6)}$ $⇔ \dfrac{36(x-6+x+6)}{(x-6)(x+6)}=\dfrac{4,5(x²-36)}{(x-6)(x+6)}$ $⇔ 4,5x²-72x-162=0$ $⇔ 4,5x²+9x-81x-162=0$ $⇔ 4,5x(x+2)-81(x+2)=0$ $⇔ (4,5x-81)(x+2)=0$ \(⇔\left[ \begin{array}{l}4,5x-81=0\\x+2=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=18\\x=-2\end{array} \right.\) Cách 2: $4,5x^2 – 72x – 162 = 0$ $Δ=(-72)²-4.4,5.(-162)$ $Δ=5184-(-2916)$ $Δ=8100$ Vì $Δ>0$ nên ta có 2 nghiệm phân biệt: $x_1=\dfrac{72+\sqrt{8100}}{2.4,5}=\dfrac{162}{9}=18$ $x_2=\dfrac{72-\sqrt{8100}}{2.4,5}=\dfrac{-18}{9}=-2$ Bình luận
`4,5x^2 – 72x + 162 = 0`
`<=>x^2 – 16x +36 = 0`
`<=>x^2+2x – 18x +36 = 0`
`<=>(x^2+2x) -( 18x +36) = 0`
`<=>x(x+2) – 18(x +2) = 0`
`<=>(x – 18)(x +2) = 0`
\(⇔\left[ \begin{array}{l}x-18=0\\x+2=0\end{array} \right.\)\(⇔\left[ \begin{array}{l}x=18\\x=-2\end{array} \right.\)
`tox={18;-2}`
Cách 1:
$\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5$
$⇔ \dfrac{36(x-6)+36(x+6)}{(x-6)(x+6)}=\dfrac{4,5(x-6)(x+6)}{(x-6)(x+6)}$
$⇔ \dfrac{36(x-6+x+6)}{(x-6)(x+6)}=\dfrac{4,5(x²-36)}{(x-6)(x+6)}$
$⇔ 4,5x²-72x-162=0$
$⇔ 4,5x²+9x-81x-162=0$
$⇔ 4,5x(x+2)-81(x+2)=0$
$⇔ (4,5x-81)(x+2)=0$
\(⇔\left[ \begin{array}{l}4,5x-81=0\\x+2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=18\\x=-2\end{array} \right.\)
Cách 2:
$4,5x^2 – 72x – 162 = 0$
$Δ=(-72)²-4.4,5.(-162)$
$Δ=5184-(-2916)$
$Δ=8100$
Vì $Δ>0$ nên ta có 2 nghiệm phân biệt:
$x_1=\dfrac{72+\sqrt{8100}}{2.4,5}=\dfrac{162}{9}=18$
$x_2=\dfrac{72-\sqrt{8100}}{2.4,5}=\dfrac{-18}{9}=-2$