Ta có: \(5^x+5^{x-1}=125\)

\(\Leftrightarrow5^{x-1}\left(5+1\right)=125\)

\(\Leftrightarrow5^{x-1}\cdot6=125\)

\(\Rightarrow5^{x-1}=\frac{125}{6}\)

\(\Leftrightarrow x-1\approx1,87\)

\(\Rightarrow x\approx2,87\)

Có vấn đề rồi đây

\(\hept{\begin{cases}2x=5y\\3x+4y=46\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{5}}\\3x+4y=46\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{3x}{\frac{3}{2}}=\frac{4y}{\frac{4}{5}}\\3x+4y=46\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{3x}{\frac{3}{2}}=\frac{4y}{\frac{4}{5}}=\frac{3x+4y}{\frac{3}{2}+\frac{4}{5}}=\frac{46}{\frac{23}{10}}=20\)

\(\frac{3x}{\frac{3}{2}}=20\Rightarrow3x=30\Rightarrow x=10\)

\(\frac{4y}{\frac{4}{5}}=20\Rightarrow4y=16\Rightarrow y=4\)

2.x=5.y   = \(\frac{X}{5}\)=\(\frac{Y}{2}\)=\(\frac{3x+4Y}{3.5+4.2}\)=\(\frac{46}{23}\)=2

\(\frac{X}{5}\)=2 => x=2.5=10

\(\frac{Y}{2}\)=2 =>y=2.2=4

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰ > 1000¹⁰ = 10³⁰

2¹⁰⁰ = 2³¹. 2⁶. 2⁶³ = 2³¹. 64 . 512⁷ < 2³¹. 125 . 625⁷ = 2³¹. 5³. 5²⁸ = 2³¹. 5³¹ = 10³¹

=> 10³⁰ < 2¹⁰⁰ < 10³¹

=> 2¹⁰⁰ có 31 chữ số

Vậy, 2¹⁰⁰ có 31 chữ số

Ta có: 

\(S=23+43+63+...+203\)

\(S=\left(10+13\right)+\left(20+23\right)+...+\left(100+103\right)\)

\(S=\left(10+20+...+100\right)+\left(13+23+...+103\right)\)

\(S=10\cdot\frac{10\cdot11}{2}+3025\)

\(S=550+3025=3575\)

Ta có: 13 + 23 + ... + 103 = 3025

=> 2 . 13 + 2 . 23 + ... + 2 . 103 = 2 . 3025

=> 26 + 46 + ... + 206 = 6050

=> ( 23 + 3 ) + ( 43 + 3 ) + ... + ( 203 + 3 ) = 6050

=>23 + 43 + ... + 203 = 6050 - 3 . 10

=>S = 6020

8x=5y => \(\frac{y}{8}=\frac{x}{5}\)

áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{y}{8}=\frac{x}{5}=\frac{y-x}{8-5}=-\frac{12}{3}=-4\)

=> \(\hept{\begin{cases}x=-4\cdot5=-20\\y=-4\cdot8=-32\end{cases}}\)

vậy.......

thank bạn nhé

a) \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

b) \(1^2+2^2+...+n^2\)

\(=1\left(2-1\right)+2\left(3-1\right)+...+n\left[\left(n+1\right)-1\right]\)

\(=1.2+2.3+...+n\left(n+1\right)-\left(1+2+...+n\right)\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{1.2.3-1.2.3+2.3.4-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=n\left(n+1\right)\left(\frac{n+2}{3}-\frac{1}{2}\right)\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

câu c đâu bạn