Ta có:

\(3^{2+n}-2^{2+n}+3^n-2^n\)

\(=3^2.3^n-2^2.2^n+3^n-2^n\)

\(=9.3^n-4.2^n+3^n-2^n\)

\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=10.3^n-5.2^n\)

\(=10.3^n-5.2.2^{n-1}\)

\(=10.3^n-10.2^{n-1}⋮10\left(đpcm\right)\) (n nguyên dương)

a^2 + b^2 = hay suy ra vậy bn

(a-b)^2>=0 với mọi a,b

<=> a^2+b^2>= 2ab

<=> (a^2 + b^2)*2 >= a^2 + b^2 + 2ab = (a+b)^2 =1

suy ra đpcm

dấu = xảy ra <=> a=b=1/2

ta có a+b=1

=> (a+b)^2=1

=> a^2+2ab+b^2=1(1)

mặt khác a+b>=\(2\sqrt{ab}=>\left(a+b\right)^2>=4ab\)

=> 1>=4ab

=>ab<=1/4 =>2ab <=1/2

thay vào(1)=> a^+b^2>=1-1/2=1/2

dấu = khi a=b=1/2

Sửa đề: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2c}{abc}+\frac{2a}{abc}+\frac{2b}{abc}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{a+b+c}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

hay \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)(đpcm)

Ta có : \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=2^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=4-2\cdot\left(\dfrac{a+b+c}{abc}\right)=4-2\cdot\dfrac{abc}{abc}=4-2\cdot1=2\)