Đáp án `+` Giải thích các bước giải `!`
`to` Phân tích đa thức thành nhân tử:
`g)`
`x^2+4x-12`
`= x^2+6x-2x-12`
`= (x^2+6x)-(2x+12)`
`= x(x+6)-2(x+6)`
`= (x-2)(x+6)`
`i)`
`x^3+8+6x(x+2)`
`= x^3+8+6x^2+12x`
`= x^3+6x^2+12x+8`
`= x^3+3. x^2. 2+3. x. 2^2+2^3`
`= (x+2)^3`
Áp dụng: `(a+b)^3 = a^3+3a^2b+3ab^2+b^3`
`k)`
`(x^2+6x+5)(x^2+10x+21)+15`
`= (x^2+5x+x+5)(x^2+7x+3x+21)+15`
`= [(x^2+5x)+(x+5)][(x^2+7x)+(3x+21)]+15`
`= [x(x+5)+(x+5)][x(x+7)+3(x+7)]+15`
`= (x+1)(x+5)(x+3)(x+7)+15`
`= [(x+1)(x+7)][(x+5)(x+3)]+15`
`= (x^2+7x+x+7)(x^2+3x+5x+15)+15`
`= (x^2+8x+7)(x^2+8x+15)+15`
Đặt `x^2+8x+7=a`
`= a(a+8)+15`
`= a^2+8a+15`
`= a^2+3a+5a+15`
`= (a^2+3a)+(5a+15)`
`= a(a+3)+5(a+3)`
`= (a+5)(a+3)`
Thay `a=x^2+8x+7` vào ta có:
`= (x^2+8x+7+5)(x^2+8x+7+3)`
`= (x^2+8x+12)(x^2+8x+10)`
`= (x^2+6x+2x+12)(x^2+8x+10)`
`= [(x^2+6x)+(2x+12)](x^2+8x+10)`
`= [x(x+6)+2(x+6)](x^2+8x+10)`
`= (x+2)(x+6)(x^2+8x+10)`
