Đáp án+Giải thích các bước giải:
Bài 2:
`a)2x(x-5)-x(3+2x)=26`
⇔ `2x^2-10x-3x-2x^2=26`
⇔ `-13x=26`
⇔ `x=-2`
Vậy `S={-2}`
`b)(12x-5)(4x+1)+(3x-7)(1-16x)=81`
⇔ `48x^2+12x-20x-5+3x-48x^2-7+112x=81`
⇔ `107x-12=81`
⇔ `107x=93`
⇔ `x=93/107`
Vậy `S={93/107}`
c)`2x^2+3(x-1)(x+1)=5x(x+1)`
⇔ `2x^2+3(x^2-1)=5x^2+5x`
⇔ `2x^2+3x^2-3-5x^2+5x=0`
⇔ `5x-3=0`
⇔ `x=3/5`
Vậy `S={3/5}`
Bài 3:
Ta có:$\left \{\begin{matrix}x=a^2-bc\\y=b^2-ac\\z=c^2-ab\end{matrix} \right.$ ⇔$\left \{\begin{matrix}ax=a^3-abc\\by=b^3-abc\\cz=c^3-abc\end{matrix} \right.$
⇔ `ax+by+cz=a^3+b^3+c^3-3abc`
`=(a+b)^3+c^3-3ab(a+b)-3abc`
`=(a+b+c)^3-3ab(a+b)-3abc-3(a+b).c.(a+b+c)`
`=(a+b+c)^3-3ab(a+b+c)-3(a+b)c(a+b+c)`
`=(a+b+c)[(a+b+c)^2-3ab-3(a+b)c]`
`=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3ac-3bc)`
`=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)`
`=(a+b+c)[(a^2-bc)+(b^2-ac)+(c^2-ab)]`
`=(a+b+c)(x+y+z)`
⇒ `(a+b+c)(x+y+z)=ax+by+cz`(đpcm)
Áp dụng:
`(a+b)^3=a^3+3a^2b+3ab^2+b^3`
`=a^3+b^3+3ab(a+b)`
`(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca`
`____________________`
Chúc bạn học tốt!!!
`#Rùa~ ~ ~`
