Ta có:
\(M=4\left(x-2\right)\left(x-1\right)\left(x+4\right)\left(x+8\right)+25x^2\)
\(=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+\left(5x\right)^2\)
Đặt \(x^2+7x-8=t\) ta có:
\(4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+\left(5x\right)^2=4t\left(t-5x\right)+\left(5x\right)^2\)
\(=4t^2-20tx+\left(5x\right)^2=\left(2t-5x\right)^2\)
Thay \(t=x^2+7x-8\) ta được:
\(\left(2t-5x\right)^2=\left[2\left(x^2+7x-8\right)-5x\right]^2=\left(2x^2+9x-16\right)^2\ge0\forall x\)
Vậy, M luôn không âm với mọi x
Ta có:
M=4(x−2)(x−1)(x+4)(x+8)+25x2M=4(x−2)(x−1)(x+4)(x+8)+25x2
=4(x2+2x−8)(x2+7x−8)+(5x)2=4(x2+2x−8)(x2+7x−8)+(5x)2
Đặt x2+7x−8=tx2+7x−8=t ta có:
4(x2+2x−8)(x2+7x−8)+(5x)2=4t(t−5x)+(5x)24(x2+2x−8)(x2+7x−8)+(5x)2=4t(t−5x)+(5x)2
=4t2−20tx+(5x)2=(2t−5x)2=4t2−20tx+(5x)2=(2t−5x)2
Thay t=x2+7x−8t=x2+7x−8 ta được:
(2t−5x)2=[2(x2+7x−8)−5x]2=(2x2+9x−16)2≥0∀x(2t−5x)2=[2(x2+7x−8)−5x]2=(2x2+9x−16)2≥0∀x
Vậy, M luôn không âm với mọi x
Ta có:
M=4(x−2)(x−1)(x+4)(x+8)+25x2M=4(x−2)(x−1)(x+4)(x+8)+25x2
=4(x2+2x−8)(x2+7x−8)+(5x)2=4(x2+2x−8)(x2+7x−8)+(5x)2
Đặt x2+7x−8=tx2+7x−8=t ta có:
4(x2+2x−8)(x2+7x−8)+(5x)2=4t(t−5x)+(5x)24(x2+2x−8)(x2+7x−8)+(5x)2=4t(t−5x)+(5x)2
=4t2−20tx+(5x)2=(2t−5x)2=4t2−20tx+(5x)2=(2t−5x)2
Thay t=x2+7x−8t=x2+7x−8 ta được:
(2t−5x)2=[2(x2+7x−8)−5x]2=(2x2+9x−16)2≥0∀x(2t−5x)2=[2(x2+7x−8)−5x]2=(2x2+9x−16)2≥0∀x
Vậy, M luôn không âm với mọi xv
