A(x)A(x)=ax2.ax2 + bx.bx + cc

= ax2ax2 +(5a+c)x(5a+c)x + cc  (vìvì bb=5a5a+cc)

=ax2ax2 + 5ax5ax + cxcx + cc

A(1)A(1)=aa + 5a5a + 2c2c

A(−3)A(−3)=9a9a - 15a15a - 3c3c + cc

=9a9a - 15a15a - 2c2c

A(1)A(1)+A(−3)A(−3)=aa + 5a5a + 2c2c+9a9a - 15a15a - 2c2c

=(aa+ 5a5a+9a9a - 15a15a)+(2c2c-2c2c)

=0

dodo đóđó A(1)A(1)=−A(−3)−A(−3)

hayhay A(1)A(1).A(−3)A(−3)=−A(−3)−A(−3).A(−3)A(−3)=-A(−3)A(−3)²≤0

hok tốt nha nhưng mik ko chắc là đúng đâu

đề bài là A(3) chứ không phải là A(-3) nhé bạn

Ta có S + 4 = \(\left(\frac{a}{b+c+d}+1\right)+\left(\frac{b}{c+d+a}+1\right)+\left(\frac{c}{a+b+d}+1\right)+\left(\frac{d}{a+b+c}+1\right)\)

\(=\frac{a+b+c+d}{b+c+d}+\frac{a+b+c+d}{a+c+d}+\frac{a+b+c+d}{a+b+d}+\frac{a+b+c+d}{b+c+d}\)

\(=\left(a+b+c+d\right)\left(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}\right)\)

\(=4000.\frac{1}{40}=100\)(a + b + c + d = 4000 ; \(\frac{1}{b+c+d}+\frac{1}{a+c+d}+\frac{1}{a+b+d}+\frac{1}{a+b+c}=\frac{1}{40}\))

=> S = 100 - 4 = 96

a) (x - 8)(x³ + 8)=  0

=> (x - 8)(x + 2)(x² - 2x + 4) = 0

=> \(\hept{\begin{cases}x-8=0\\x+2=0\\x^2-2x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)

Vì x² - 2x + 4 = (x² - 2.x.1 + 1²)+3 \(\ge\)3 > 0

Vậy x \(\in\){8;-2}

b) (4x - 3) - (x + 5) = 3 .(10 - x)

=> 4x - 3 - x - 5 = 30  - 3x

=> 4x - 3 - x - 5 - 30 + 3x = 0

=> 6x - 38 = 0

=> 3x - 19 = 0

=> x = 19/3

Vậy x = 19/3

f(x) = x⁶ - 2009x⁵ + 2009x⁴ - 2009x³ + 2009x² - 2009x + 2011

x = 2008 => 2009 = x + 1

=> f(2008) = f(x+1) = x⁶ - (x+1)x⁵ + (x+1)x⁴ - (x+1)x³ + (x+1)x² - (x+1)x + 2011

= x⁶ - x⁶ - x⁵ + x⁵ + x⁴ - x⁴ - x³ + x³ + x² - x² - x + 2011

= -x + 2011 = -2008 + 2011 = 3

Vậy ...

đỏ đi mà làm người