Ta có: \(A=2+2^2+2^3+...+2^{100}\)

\(2A=2^2+2^3+2^4+...+2^{101}\)

\(2A-A=2^{101}-2\)

Hay \(A=2^{101}-2\)

Vậy \(A=2^{101}-2\)

_Học tốt_

bạn trả lời quá chậm

a) VP = (a+b)³ - 3ab(a+b)

         =[a³ + b³ + 3ab(a+b)] - 3ab(a+b)

        = a³ + b³ = VT

b) 

a³+b³+c³−3abc

=(a+b)³+c³−3a²b−3ab²−3abc

=(a+b+c)³[(a+b)²−(a+b)c+c²]−3ab(a+b)−3abc

=(a+b+c)(a²+b²+2ab−ac−bc+c²)−3ab(a+b+c)

=(a+b+c)(a²+b²+2ab−ac−bc+c²−3ab)

=(a+b+c)(a²+b²+c²-ab-bc-ca) (đpcm)

nhớ đúng cho mk nha !!!!!

Chứng minh tương đương: 

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

<=> \(\frac{a^2+b^2+2}{1+a^2+b^2+a^2b^2}\ge\frac{2}{1+ab}\)

<=> \(\left(1+ab\right)\left(a^2+b^2+2\right)\ge2\left(1+a^2+b^2+a^2b^2\right)\)

<=> \(a^2+b^2+2+a^3b+ab^3\ge2+2a^2+2b^2+2a^2b^2\)

<=> \(a^3b+ab^3-2a^2b^2\ge a^2+b^2-2ab\)

<=> \(ab\left(a^2+b^2-2ab\right)\ge a^2+b^2-2ab\)

<=> \(ab\left(a-b\right)^2\ge\left(a-b\right)^2\)

<=> \(\left(a-b\right)^2\left(ab-1\right)\ge0\) luôn đúng vì a>1; b>1

Dấu "=" xảy ra <=> a - b = 0 <=> a = b.

2^x+1-15=17

2^x+1=17+15

2^x+1=32

2^x+1=2⁵

=> x+1=5

     x=5-1

    x=4

Vậy x=4.

b) \(\hept{\begin{cases}\frac{5}{2x+6}=\frac{5}{2\left(x+3\right)}\\\frac{3}{x^2-9}=\frac{3}{\left(x+3\right)\left(x-3\right)}\end{cases}}\)

\(\Rightarrow MTC=2\left(x+3\right)\left(x-3\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{5}{2\left(x+3\right)}=\frac{5\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}\\\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{6}{2\left(x-2\right)\left(x+3\right)}\end{cases}}\)

CÒn lại tương tự nhé !