3 x x² - 6 x x

=3x² - 6x + 3 - 3

=3 ( x² -2x + 1 ) +3

=> 3 ( x -1)² > -3

Vậy GTNN của D là -3

\(D=3\times\left(x^2-2x\right)\)

\(D=3\times\left(x^2-2x+1-1\right)\)

\(D=3\times\left[\left(x-1\right)^2-1\right]\)

\(D=3\times\left(x-1\right)^2-3\)

Nhận xét: \(3\times\left(x-1\right)^2\ge0\)

                \(=>3\times\left(x-1\right)^2-3\ge-3\)

                \(=>D\ge-3\)

Vậy D đạt GTNN tại D=-3 <=> x=1

Ta có:

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

\(<=>\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(<=>\frac{a+b-c+2c}{c}=\frac{b+c-a+2a}{a}=\frac{c+a-b+2b}{b}\)

\(<=>\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

\(<=>\left(a+b+c\right)\times\frac{1}{c}=\left(a+b+c\right)\times\frac{1}{a}=\left(a+b+c\right)\times\frac{1}{b}\)

\(<=>\frac{1}{a}=\frac{1}{c}=\frac{1}{b}<=>a=c=b\)

Vậy:

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

\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2^3=8\)

Vậy A=8

Xét vế trái: A\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

=>\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

=>\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)

=>đpcm         (VP là vế phải)

làm ơn giúp mình!

đề là sao vậy