Phương pháp đổi biến

T

taituchuphan

M=(x-1)(x-2)(x+4)(x+8)+25x^2
M=(x^2 -3x+2)(x^2+12x+32)+25x^2
M=((x^2-2*3/2 x+9/4)-1/4))((x^2 +2*6x+36)-4))+25x^2
vì (x-3/2)^2-1/4>=1/4
va(x+6)^2-4>=4
vi vay M=1/4*4 +25x^2>0
vi x^2>0
vay gia tri cua m khong am
 
T

thaonguyenkmhd

Ta có $M = 4(x-2)(x-1)(x+4)(x+8)+25x^2\\ \Longrightarrow M =4[(x-1)(x+8)][(x-2)(x+4)]+25x^2\\ \Longrightarrow M =4(x^2+7x-8)(x^2+2x-8)+(5x)^2$

Đặt $x^2+7x-8=y \Longrightarrow M=4y(y-5x)+(5x)^2=(2y)^2-20xy+(5x)^2=(2y-5x)^2$

Thay $y=x^2+7x-8 \Longrightarrow M$$ =[2(x^2+7x-8)-5x]^2\\=(2x^2+14x-16-5x)^2\\=(2x^2+9x-16)^2 $

Vạy M ko âm với mọi x
 
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