Đáp án: + Giải thích các bước giải:
Bài `1:`
` a) 16x^2-8x+1-3(4x-1)`
`=(4x-1)^2-3(4x-1)`
`=(4x-1)(4x-1-3)`
`=(4x-1)(4x-4)`
`=4(4x-1)(x-1)`
`b) 27x^3+8`
`=(3x)^3+2^3`
`=(3x+2)(9x^2-6x+4)`
`c) -16x^4y^6-24x^5y^5-9x^6y^4`
`=-x^4y^4(16y^2+24xy+9x^2)`
`=-x^4y^4(4y+3x)^2`
`d) (ax+by)^2-(ay+bx)^2`
`=(ax+by-ay-bx)(ax+by+ay+bx)`
`=[a(x-y)-b(x-y)][a(x+y)+b(x+y)]`
`=(a-b)(x-y)(a+b)(x+y)`
Bài `2:`
`a) (a^2+b^2-5)^2-2(ab+2)^2`
`=(a^2+b^2-5)^2-(\sqrt{2}ab+2\sqrt{2})^2`
`=(a^2+b^2-5-\sqrt{2}ab-2\sqrt{2})(a^2+b^2-5+\sqrt{2}ab+2\sqrt{2})`
`b) (4a^2-3a-18)^2-(4a^2+3a)^2`
`=(4a^2-3a-18+4a^2+3a)(4a^2-3a-18-4a^2-3a)`
`=(8a^2-18)(-6a-18)`
`=12(4a^2-9)(-a-3)`
`=12(2a-3)(2a+3)(-a-3)`
`c)-(x+2)+3(a^2-4)`
`=-(x+2)+3(x-2)(x+2)`
`=(x+2)[1-3(x-2)]`
`=(x+2)(1-3x+6)`
`=(x+2)(7-3x)`
`d) 125a^3 -27b^3`
`=(5a)^3 -(3b)^3`
`=(5a-3b)(5a+3b)`
Bài `3: `
`a)104^2-16`
`=104^2 -4^2`
`=(104-4)(104+4)`
`=100 .108`
`=10800`
`b)9^8 .2^8 -(18^4-1)(18^4+1)`
`=(9.2)^8-(18^4)^2+1`
`=18^8 -18^8+1`
`=1`
`d) 42^3 -6.42^2+12.42 -8`
`=42^3 -3.2.42^2+3.2^2. 42^2-2^3`
`=(42-2)^3`
`=40^3`
`=64000`
Bài `4:`
`a)x(x-2012)-2013x+2012.2013=0`
`⇔x(x-2012)-2013(x-2012)=0`
`⇔(x-2013)(x-2012)=0`
`⇔`\(\left[ \begin{array}{l}x-2013=0\\x-2012=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2013\\x=2012\end{array} \right.\)
Vậy `x=2013` hoặc `x=2012`
`b)(x-1)^3+1+3x(x-4)=0`
`⇔x^3-3x^2+3x-1+1+3x^2-12x=0`
`⇔x^3-9x=0`
`⇔x(x^2-9)=0`
`⇔x(x-3)(x+3)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-3=0\\x+3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=±3\end{array} \right.\)
Vậy `x=0` hoặc `x=±3`
`c) (x+4)^2-16=0`
`⇔(x+4)^2-4^2=0`
`⇔(x+4+4)(x+4-4)=0`
`⇔x(x+8)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x=x+8=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.\)
Vậy `x=0` hoặc `x=-8`
`d)12x-x^2-36=0`
`⇔-(x^2-12x+36)=0`
`⇔x^2-12x+36=0`
`⇔x^2-2.x.6+6^2=0`
`⇔(x-6)^2=0`
`⇔x-6=0`
`⇔x=6`
Vậy `x=6`
Bài `5: `
`a) A=-2x^2-8x-10` (mk nghĩ là đề ntn)
`A=-2(x^2+4x+5)`
`A=-2(x^2+4x+4)-2`
`A=-2(x+2)^2-2≥-2`
Dấu `=` xảy ra khi `x+2=0 ⇔x=-2`
Vậy `A_(min)=-2` khi `x=-2`
`b) B=9x-3x^2`
`B=-3(x^2-3x)`
`B=-3(x^2-2.3/2x+9/4-9/4)`
`B=-3(x^2-2.3/2x+9/4)+27/4`
`B=-3(x-3/2)^2+27/4≤27/4`
Dấu `=` xảy ra khi `x-3/2=0⇔x=3/2`
Vậy `B_(max)=27/4` khi `x=3/2`
