Đặt \(A=6^2+6^4+6^6+...+6^{98}+6^{100}\)

Ta có: \(A=6^2+6^4+6^6+...+6^{98}+6^{100}\)

\(\Leftrightarrow36A=6^4+6^6+...6^{100}+6^{102}\)

\(\Leftrightarrow A-36A=6^2+6^4+6^6+...6^{98}+6^{100}-6^4-6^6-...-6^{100}-6^{102}\)

\(\Leftrightarrow-35\cdot A=6^2-6^{102}\)

\(\Leftrightarrow A=\dfrac{6^{102}-6^2}{35}\)

a, 6¹⁰⁰ - 1 = (6 . 6 . 6 ..... 6) - 1 = [(...6) . (...6) . (...6) ..... (...6)] - 1 = (...6) - 1 = ...5 \(⋮\) 5

b, 21²⁰ - 11¹⁰ = (21 . 21 . 21 . 21 . 21..... 21) - (11 . 11 . 11 . 11 ..... 11) = [(...1) . (...1) . (...1) . (...1).....(...1)] - [(...1) . (...1) . (...1) . (...1).....(...1)] = (...1) - (...1) = ....0 \(⋮\) 2; \(⋮\) 5

a/ \(2014^x+80=3^y\)

- Với \(x=0\Rightarrow2014^0+80+3^y\Leftrightarrow81=3^y\Leftrightarrow3^4=3^y\Rightarrow y=4\)

- Với \(x>0\) ta có \(2014\) chẵn \(\Rightarrow2014^x\) chẵn, lại có \(80\) chẵn \(\Rightarrow\) vế trái là một số chẵn

Mà \(3^y\) luôn lẻ với mọi \(y\in N\Rightarrow\) vế phải là số lẻ

Vế trái chẵn, vế phải lẻ \(\Rightarrow\) vô nghiệm

Vậy \(\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\) là cặp nghiệm tự nhiên duy nhất

b/

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{8}+...+\frac{1}{16}+...+\frac{1}{32}+...+\frac{1}{63}\)

\(A< 1+2.\frac{1}{2}+4.\frac{1}{4}+8.\frac{1}{8}+16.\frac{1}{16}+32.\frac{1}{32}\)

\(A< 1+1+1+1+1+1=6\) (đpcm)

a) (-2).(-6)^2+3.(-2)^3

=(-2).36-24=-96

b) (-6).4.(-7).25

=42.100=4200

c) (-85)-105+62

=-190+62=-128

d) 24.(-15)+(-15).75-15

=(-15).(24+75)-15

=(-15).99-15=-1500

câu a mình ghi nhầm 

a) (-2).(-6)^2+3.(-2)^3-62

a, \(\dfrac{62}{7}.x=\dfrac{29}{90}.\dfrac{3}{56}\)

\(\dfrac{62}{7}.x=\dfrac{29}{1680}\)

\(x=\dfrac{29}{1680}:\dfrac{62}{7}\)

\(x=\dfrac{29}{14880}\)

b, \(\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{1}{7}\)

\(\dfrac{1}{5}:x=\dfrac{2}{35}\)

\(x=\dfrac{1}{5}:\dfrac{2}{35}\)

\(x=\dfrac{7}{2}\)

c, \(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{13}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\dfrac{23}{12}=\dfrac{7}{46}\)

\(\left(x+\dfrac{-1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\)

\(\left(x+\dfrac{-1}{12}\right)=\dfrac{7}{46}.\dfrac{23}{12}\)

\(x+\dfrac{-1}{12}=\dfrac{7}{24}\)

\(x=\dfrac{7}{24}-\dfrac{-1}{12}\)

\(x=\dfrac{3}{8}\)

lay ong di qua lay ba di lai cho xin may tick

lay ong di qua lay ba di lai cho xin may tick