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\(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

Suy ra \(A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

uk ok chắc có tên giống nên nhầm á bạn

ủa là sao??

\(-2^{300}=\left(-2^3\right)^{100}=-8^{100}\)

\(-3^{200}=\left(-3^2\right)^{100}=-9^{100}\)

Vì \(-8^{100}>-9^{100}\)nên \(-2^{300}>-3^{200}\)

TL

\(-2^{300}=\left(-2^3\right)^{100}\)=\(-8^{100}\)

\(-3^{200}=\left(-3^2\right)^{100}\)= \(-9^{100}\)

\(-8^{100}>-9^{100}\)(do cái này là số nguyên âm nên cái có GTTĐ lớn hơn sẽ nhỏ hơn)

=>\(-2^{300}>-3^{200}\)

3^2x=81

=>3^2x=3⁴

=>2x=4

=>x=4:2=2

\(3^{2\text{x}}=81\)

\(3^{2\text{x}}=3^4\)

\(\Rightarrow\text{ 2x}=4\)

\(\text{x}=4\div2\)

\(\text{x}=2\)

Vậy \(\text{x}=2\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

TL :

\(m=3\)

\(n=0\)

Vì 

\(5^3+2^0=5.5.5+1=125+1=126\)

HT

TL :

5^m + 2ⁿ = 126

5^m + 2ⁿ = 125 + 1

5^m = 5.5.5.

2ⁿ = 1

5.5.5  = 125

2⁰ =1

vậy m = 3 ; n =0