Giải pt à bạn:P?
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)^3=0\)
\(\Leftrightarrow x^3+4^3-\left(x^3-8-6x^2+12x\right)=0\)
\(\Leftrightarrow x^3+4^3-x^3+8+6x^2-12x=0\)
\(\Leftrightarrow72+6x^2-12x=0\Leftrightarrow6\left(x^2-2x+12\right)=0\Leftrightarrow x^2-2x+12=0\)
Ta lại có: \(x^2-2x+12=x^2-2x+1+11=\left(x-1\right)^2+11\ge11>0\ne0\)
=> Pt vô nghiệm.
(x+4).(x2-4x+16)-(x-2)3
= x3 - 4x2 + 16x + 4x2 - 16x + 64 - (x3 -4x2 + 8x - 4)
= x3 - 4x2 + 16x + 4x2 - 16x + 64 - x3 +4x2 - 8x + 4
= (x3 - x3 ) + (- 4x2 + 4x2 +4x2 ) +(16x - 16x - 8x)+ (64 + 4)
= 4x2 - 8x + 68
= 2(2x2 - 4x + 34)
Ủa đề là rút gọn biểu thức hả, lần sau ghi đề vào bạn nhé:Đ
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)^3=x^3+4^3-\left(x^3-8-6x^2+12x\right)=x^3+4^3-x^3+8+6x^2-12x=72+6x^2-12x.\)
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)^3\)
=\(x^3+4^4-\left(x^3-6x^2+12x-8\right)\)
=\(x^3+64-\left(x^3-6x^2+12x-8\right)\)
=\(6x^2-12x+72\)
(x+4).(x2-4x+16)-(x-2)3
= x3 - 4x2 + 16x + 4x2 - 16x + 64 - (x3 -6x2 + 12x - 8)
= x3 - 4x2 + 16x + 4x2 - 16x + 64 - x3 +6x2 - 12x + 8
= (x3 - x3 ) + (- 4x2 + 4x2 +6x2 ) +(16x - 16x - 12x)+ (64 + 8)
= 6x2 - 12x + 512
(x+4).(x2-4x+16)-(x3-3x2.2+3x.4-8)
(x+4).(x2-4x+16)-(x3-6x2+12x-8)
X3-4x2+16x+4x2-16x+64-x3+6x2-12x+8
6x2-12x+72