Toán Giải phương trình: (4x – 3)^3 + (3x – 2)^3 = (7x – 5)^3 06/10/2021 By Gianna Giải phương trình: (4x – 3)^3 + (3x – 2)^3 = (7x – 5)^3
(4x – 3)^3 + (3x – 2)^3 = (7x – 5)^3 ⇔ (4x – 3 + 3x – 2).[(4x-3)^2-(4x-3)(3x-2)+(3x-2)^2] = (7x – 5)^3 ⇔ (7x-5).[16x^2-24x+9-(12x^2-8x-9x+6)+9x^2-12x+4) = (7x – 5)^3 ⇔ (7x-5).(16x^2-24x+9- 12x^2 + 17x – 6 +9x^2-12x+4) = (7x – 5)^3 ⇔ (7x – 5).(13x^2-19x+7) – (7x – 5)^3 = 0 ⇔ (7x – 5).[ 13x^2-19x+7 – (7x – 5)^2 ] = 0 ⇔ (7x – 5).( 13x^2-19x+7 – 49x^2 – 70x +25 ) = 0 ⇔ (7x – 5).(-36x^2 + 51x – 18 ) = 0 ⇔ (7x – 5).[ 3(12x^2 – 17x + 6 )] = 0 ⇔ (7x – 5).[ 3(12x^2 – 8x – 9x + 6 )] = 0 ⇔ (7x-5).{ 3[4x(3x-2)-3(3x-2)]} = 0 ⇔ (7x-5).{ 3[(3x-2)(4x-3)]} = 0 ⇔ (7x-5).(9x-6).(4x-3) = 0 ⇔ 7x-5 = 0 hoặc 9x-6=0 hoặc 4x-3=0 ⇒ x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$ Vậy x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$ Trả lời
`(4x-3)^3+(3x-2)^3=(7x-5)^3` `<=>(4x-3+3x-2)[(4x-3)^2-(4x-3)(3x+2)+(3x-2)^2=7x-5^3` `<=>(7x-5)[(16x-24x+9+12x^2+8x+9x-6+9x^2-12x+4)-(7x-5)^3]=0` `<=>(7x-5)(13x^2-19x+7-(7x-5)^2=0` `<=>(7x-5)(13x^2-19x+7-49x^2+70x-25)=0` `<=>(7x-5)(-36x^2+51x-18)=0` `<=>`\(\left[ \begin{array}{l}7x-5=0\\36x^2-51x+18=0\end{array} \right.\) $↔️\left[ \begin{array}{l}x=\dfrac{5}{7}\\36x^2-51x+18=0\end{array} \right.$ `<=>36x^2-517+18=0` `Δ=51^2-4.36.18=9>0=>` \(\left[ \begin{array}{l}x=\dfrac{51-3}{72}=\dfrac{2}{3}\\x=\dfrac{51+3}{72}=\dfrac{3}{4}\end{array} \right.\) Vậy `pt` có nghiệm `{5/7;2/3;3/4}` Trả lời
(4x – 3)^3 + (3x – 2)^3 = (7x – 5)^3
⇔ (4x – 3 + 3x – 2).[(4x-3)^2-(4x-3)(3x-2)+(3x-2)^2] = (7x – 5)^3
⇔ (7x-5).[16x^2-24x+9-(12x^2-8x-9x+6)+9x^2-12x+4) = (7x – 5)^3
⇔ (7x-5).(16x^2-24x+9- 12x^2 + 17x – 6 +9x^2-12x+4) = (7x – 5)^3
⇔ (7x – 5).(13x^2-19x+7) – (7x – 5)^3 = 0
⇔ (7x – 5).[ 13x^2-19x+7 – (7x – 5)^2 ] = 0
⇔ (7x – 5).( 13x^2-19x+7 – 49x^2 – 70x +25 ) = 0
⇔ (7x – 5).(-36x^2 + 51x – 18 ) = 0
⇔ (7x – 5).[ 3(12x^2 – 17x + 6 )] = 0
⇔ (7x – 5).[ 3(12x^2 – 8x – 9x + 6 )] = 0
⇔ (7x-5).{ 3[4x(3x-2)-3(3x-2)]} = 0
⇔ (7x-5).{ 3[(3x-2)(4x-3)]} = 0
⇔ (7x-5).(9x-6).(4x-3) = 0
⇔ 7x-5 = 0 hoặc 9x-6=0 hoặc 4x-3=0
⇒ x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$
Vậy x = $\frac{5}{7}$ hoặc x= $\frac{2}{3}$ hoặc x = $\frac{3}{4}$
`(4x-3)^3+(3x-2)^3=(7x-5)^3`
`<=>(4x-3+3x-2)[(4x-3)^2-(4x-3)(3x+2)+(3x-2)^2=7x-5^3`
`<=>(7x-5)[(16x-24x+9+12x^2+8x+9x-6+9x^2-12x+4)-(7x-5)^3]=0`
`<=>(7x-5)(13x^2-19x+7-(7x-5)^2=0`
`<=>(7x-5)(13x^2-19x+7-49x^2+70x-25)=0`
`<=>(7x-5)(-36x^2+51x-18)=0`
`<=>`\(\left[ \begin{array}{l}7x-5=0\\36x^2-51x+18=0\end{array} \right.\) $↔️\left[ \begin{array}{l}x=\dfrac{5}{7}\\36x^2-51x+18=0\end{array} \right.$
`<=>36x^2-517+18=0`
`Δ=51^2-4.36.18=9>0=>` \(\left[ \begin{array}{l}x=\dfrac{51-3}{72}=\dfrac{2}{3}\\x=\dfrac{51+3}{72}=\dfrac{3}{4}\end{array} \right.\)
Vậy `pt` có nghiệm `{5/7;2/3;3/4}`