Bài 1: Tìm giá trị nhỏ nhất của:
\(A=\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x+7\right)\)
\(B=9x^2-6x+2\)
\(C=x^2+x+1\)
\(D=2x^2+2x+1\)
Bài 2: Tính \(a^4+b^4+c^4\) biết:
a, \(a+b+c=0\)và \(a^2+b^2+c^2=2\)
b, \(a+b+c=0\)và \(a^2+b^2+c^2=1\)
2/
a,Ta có: a+b+c=0
<=>(a+b+c)2=0
<=>a2+b2+c2+2(ab+bc+ca)=0
<=>2+2(ab+bc+ca)=0
<=>ab+bc+ca=\(\frac{-2}{2}=-1\)
<=>(ab+bc+ca)2=1
<=>a2b2+b2c2+c2a2+2abc(a+b+c)=1
<=>a2b2+b2c2+c2a2=1 (vì a+b+c=0)
Lại có: a2+b2+c2=2
<=>(a2+b2+c2)2=4
<=>a4+b4+c4+2(a2b2+b2c2+c2a2)=4
<=>a4+b4+c4+2=4 (vì a2b2+b2c2+c2a2=1)
<=>a4+b4+c4=2
b, tương tự a
1/
b, \(B=9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1\)
Vì \(\left(3x-1\right)^2\ge0\Rightarrow B=\left(3x-1\right)^2+1\ge1\)
Dấu "=" xảy ra khi x=1/3
Vậy Bmin = 1 khi x = 1/3
c,\(C=x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra khi x=-1/2
Vậy...
d, \(D=2x^2+2x+1=2\left(x^2+x+\frac{1}{2}\right)=2\left(x^2+x+\frac{1}{4}+\frac{1}{4}\right)=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\)
Vì \(2\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow D=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
Dấu "=" xảy ra khi x=-1/2
Vậy...