\(99^{20}=99^{2\cdot10}=\left(99^2\right)^{10}=9801^{10}\)

Vì \(9801^{10}>999^{10}\)

Nên \(99^{20}>999^{10}\)

99²⁰ = 99^2.10 = (99²)¹⁰ = 9801¹⁰

Có 9801 > 999

=> 9801¹⁰ > 999¹⁰

=> 99²⁰ > 999¹⁰

Trả lời:

a)\(\frac{18}{87}>\frac{17}{98}\)

b)\(\frac{999}{100}>\frac{998}{999}\)

Hok tốt!

bn ơi giải thích rõ ràng hộ mình đc ko

Vì \(\frac{18}{87}>\frac{17}{87}\)

Mà \(\frac{17}{87}>\frac{17}{98}\)(  khi so sánh hai p/s có tử giống nhau có mẫu nào lớn hơn thì bé hơn và ngược lại )

Dựa vào tính chất bắc cầu 

=> \(\frac{18}{87}>\frac{17}{87}>\frac{17}{98}\)

=>\(\frac{18}{87}>\frac{17}{98}\)

Vậy ...

b,.....

hay chi cho to voi 

1024¹⁰ = (2¹⁰)¹⁰ = 2¹⁰⁰

Vì 2¹⁰⁰ < 10¹⁰⁰ nên 1024¹⁰ < 10¹⁰⁰

cam on ban

a) \(1.2+2.3+3.4+...+19.20\)

\(=\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)

\(=3080\)

b) \(9+99+999+...+999...9\left(100so9\right)\)

\(\)\(=\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+...+\left(1000...0-1\right)\left(99so0\right)\)

\(=\left(10+10^2+10^3+...10^{99}\right)+\left(-1\right).100\)

\(=\left(1+10+10^2+10^3+...10^{99}\right)+\left(-1\right).101\)

\(=\dfrac{10^{99+1}-1}{99-1}-101\)

\(=\dfrac{10^{100}-1}{98}-101\)

\(=\dfrac{10^{100}-9899}{98}\)

c) \(999.9x222...2\) (100 số 9; 100 số 2)

\(9x2=18\)

\(99x22=2178\)

\(999x222=\text{221778}\)

\(9999x2222=22217778\)

\(99999x22222=2222177778\)

\(.........\)

Theo quy luật trên ta có 100 số 9 nhân 100 số 2:

\(999.9x222...2=222...21777...78\) (99 sô 2; 1 số 1; 99 số 7; 1 số 8)

A, 1.2 + 2. 3 + 3. 4 + ....+ 19 . 20 

⇒\(\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)

⇒3080 

vậy kết quả câu a, là 3080 

A = 10³⁰ =(10³)¹⁰ = 1000¹⁰

B = (2¹⁰)¹⁰ =1024¹⁰

Vì 1000 < 1024

=> A <B

\(2^{1000}=\left(2^{10}\right)^{100}=1024^{100}>10^{100}\)

10^100 = (2 x 5) ^100 = 2^100 x 5 ^100 

2^ 1000 = 2^ 100x 10 = 2^100 x 2^ 10 

vì 2 ^10 < 5^ 100 

=> ...............

=> 10^100 < 2^1000

sorry cac ban nha

minh chi co mot nick nen chi h dc 1 ban

\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)

\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)

\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)

\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)

\(=\dfrac{7}{2\left(1+6\right)}\)

\(=\dfrac{7}{2.7}\)

\(=\dfrac{1}{2}\)

a) \(5^{20}và2550^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)

=> \(5^{20}< 2550^{10}\)

b) \(999^{10}và999999^5\)

\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)

=> \(999^{10}< 999999^5\)

c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)

\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)

\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)

\(\dfrac{-1}{5}=\dfrac{-3}{15}\)

\(\dfrac{-1}{3}=\dfrac{-5}{15}\)

=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)

=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)