\(S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{51}}\)

\(\Rightarrow\)\(2S=2+1+\frac{1}{2}+....+\frac{1}{2^{50}}\)

\(\Rightarrow\)\(2S-S=2-\frac{1}{5^{51}}\)

\(\Rightarrow\)\(S=2-\frac{1}{5^{51}}\)

Ta có:

27⁵ = (3³)⁵ = 3¹⁵

243³ = (3⁵)³ = 3¹⁵

Vì 3¹⁵ = 3¹⁵ => 27⁵ = 243³

Mà 243³ < 245³ (vì 243 < 245)

=> 27⁵ < 245³

Vậy 27⁵ < 245³

27⁵=245³

1/ 

Từ \(a-b=2\left(a+b\right)\Rightarrow a-b=2a+2b\Rightarrow a-2a=2b+b\Rightarrow-a=3b\Rightarrow a=-3b\)

\(\Rightarrow\frac{a}{b}=\frac{-3b}{b}=-3\)

\(\Rightarrow\hept{\begin{cases}a-b=-3\\2\left(a+b\right)=-3\end{cases}\Rightarrow\hept{\begin{cases}a-b=-3\\a+b=-\frac{3}{2}\end{cases}}}\)

\(\Rightarrow a-b+a+b=-3-\frac{3}{2}\Rightarrow2a=\frac{-9}{2}\Rightarrow a=\frac{-9}{4}\)

Có: \(a-b=-3\Rightarrow b=a+3\Rightarrow b=\frac{-9}{4}+3=\frac{3}{4}\)

Vậy a=-9/4,b=3/4

2/ Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)

Ta có: \(\frac{bx-ay}{a}=\frac{bak-abk}{a}=0\left(1\right)\)

\(\frac{cx-az}{y}=\frac{cak-ack}{y}=0\left(2\right)\)

\(\frac{ay-bx}{c}=\frac{abk-bak}{c}=0\left(3\right)\)

Từ (1),(2),(3) => đpcm

8^12 =(2^3)^12 = 2^36 

25^19 =(5^2)^19 = 5^38 = 5^36 .5^2 = 5^36 .25

Ta có: a.b = 2^36 .5^36 .25

                = 10^36 .25

                = 2500...000

                      36 c/s 0

Vậy a có 38 chữ số.

Chúc bạn học tốt.

\(A=75\left(4^{2004}+...+4+1\right)+25\)

\(=25\left(4-1\right)\left(4^{2004}+...+4+1\right)+25\)

\(=25\left[4\left(4^{2004}+...+4+1\right)-\left(4^{2004}+...+4+1\right)\right]+25\)

\(=25\left[\left(4+4^2+...+4^{2005}\right)-\left(1+4+...+4^{2004}\right)\right]+25\)

\(=25\left(4^{2005}-1\right)+25\)

\(=25.4^{2005}-25+25\)

\(=100.4^{2004}⋮100\)

ồ cuk khó nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

a, \(A=\left|2x-5\right|+\left|2x-12\right|=\left|2x-5\right|+\left|12-2x\right|\ge\left|2x-5+12-2x\right|=7\)

Dấu "=" xảy ra khi \(\left(2x-5\right)\left(12-2x\right)\ge0\Leftrightarrow\frac{5}{2}\le x\le6\)

Vậy Amin=7 khi 5/2 <= x <= 6

b, \(B=\left|3x+6\right|+\left|3x-8\right|=\left|3x+6\right|+\left|8-3x\right|\ge\left|3x+6+8-3x\right|=14\)

Dấu "=" xảy ra khi \(\left(3x+6\right)\left(8-3x\right)\ge0\Leftrightarrow-2\le x\le\frac{8}{3}\)

Vậy...

c, \(C=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=\left(\left|x-1\right|+\left|3-x\right|\right)+\left(\left|x-2\right|+\left|4-x\right|\right)\ge\left|x-1+3-x\right|+\left|x-2+4-x\right|=2+2=4\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left(x-2\right)\left(4-x\right)\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}1\le x\le3\\2\le x\le4\end{cases}\Leftrightarrow}2\le x\le3}\)

Vậy...

\(\frac{x}{4}=\frac{3}{2}\)

\(\Rightarrow\frac{x}{4}=\frac{6}{4}\)

\(\Rightarrow x=6\)

vậy_

b)\(\frac{2}{x}=\frac{x}{8}\)

\(\Rightarrow x^2=2\cdot8\)

\(x^2=16\Rightarrow x=4\)

c) \(\frac{x+3}{4}=\frac{5}{3}\)

\(3\left(x+3\right)=4\cdot5\)

\(3x+9=20\)

\(3x=11\)

\(x=\frac{11}{3}\)