giải phương trình:$\frac{x^2}{3}$ + $\frac{48}{x^2}$ = 5($\frac{x}{3}$ +$\frac{4}{x}$) 13/10/2021 Bởi Harper giải phương trình:$\frac{x^2}{3}$ + $\frac{48}{x^2}$ = 5($\frac{x}{3}$ +$\frac{4}{x}$)
Đáp án: `S={2;6}` Giải thích các bước giải: `(x^2)/3+(48)/(x^2)=5(x/3+4/x)(ĐKXĐ:x\ne 0)` `<=>(x^4+144)/(3x^2)=5(x(x^2+12)/(3x^2))` `<=>x^4+144=5x^3+60x` `<=>x^4-5x^3-60x+144=0` `<=>x^4-2x^3-3x^3+6x^2-6x^2+12x-72x+144=0` `<=>x^3(x-2)-3x^2(x-2)-6x(x-2)-72(x-2)=0` `<=>(x-2)(x^3-3x^2-6x-72)=0` `<=>(x-2)(x^3-6x^2+3x^2-18x+12x-72)=0` `<=>(x-2)[x^2(x-6)+3x(x-6)+12(x-6)]=0` `<=>(x-2)(x-6)(x^2+3x+12)=0` `<=>`\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\)`(Vì x^2+3x+12 \ne0)` `<=>`\(\left[ \begin{array}{l}x=2(t/m)\\x=6(t/m)\end{array} \right.\) Vậy `S={2;6}` Bình luận
`(x^2)/3+(48)/(x^2)=5(x/3+4/x)` `⇔(x^4)/(3x^2)+(144)/(3x^2)=5(x^3/(3x^2)+(12x)/(3x^2))` `⇔x^4+144=5x^3+60x` `⇔x^4+144-5x^3-60x=0` `⇔x^3(x-2) -3x^2(x-2)-6x(x-2)-72(x-2)=0` `⇔(x-2)(x^3-3x^2-6x-72)=0` `⇔(x-2)[x^2(x-6) +3x(x-6) +12(x-6)]=0` `⇔(x-2)(x-6)(x^2+3x+12)=0` `⇔(x-2)(x-6)=0 ( vì x^2+3x+12>0)“ ⇔\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\) Bình luận
Đáp án:
`S={2;6}`
Giải thích các bước giải:
`(x^2)/3+(48)/(x^2)=5(x/3+4/x)(ĐKXĐ:x\ne 0)`
`<=>(x^4+144)/(3x^2)=5(x(x^2+12)/(3x^2))`
`<=>x^4+144=5x^3+60x`
`<=>x^4-5x^3-60x+144=0`
`<=>x^4-2x^3-3x^3+6x^2-6x^2+12x-72x+144=0`
`<=>x^3(x-2)-3x^2(x-2)-6x(x-2)-72(x-2)=0`
`<=>(x-2)(x^3-3x^2-6x-72)=0`
`<=>(x-2)(x^3-6x^2+3x^2-18x+12x-72)=0`
`<=>(x-2)[x^2(x-6)+3x(x-6)+12(x-6)]=0`
`<=>(x-2)(x-6)(x^2+3x+12)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\)`(Vì x^2+3x+12 \ne0)`
`<=>`\(\left[ \begin{array}{l}x=2(t/m)\\x=6(t/m)\end{array} \right.\)
Vậy `S={2;6}`
`(x^2)/3+(48)/(x^2)=5(x/3+4/x)`
`⇔(x^4)/(3x^2)+(144)/(3x^2)=5(x^3/(3x^2)+(12x)/(3x^2))`
`⇔x^4+144=5x^3+60x`
`⇔x^4+144-5x^3-60x=0`
`⇔x^3(x-2) -3x^2(x-2)-6x(x-2)-72(x-2)=0`
`⇔(x-2)(x^3-3x^2-6x-72)=0`
`⇔(x-2)[x^2(x-6) +3x(x-6) +12(x-6)]=0`
`⇔(x-2)(x-6)(x^2+3x+12)=0`
`⇔(x-2)(x-6)=0 ( vì x^2+3x+12>0)“
⇔\(\left[ \begin{array}{l}x-2=0\\x-6=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=6\end{array} \right.\)