\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{2^3\times3^3+3\times2^2\times3^2+3^3}{-13}\)

                                    \(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)

                                     \(=\frac{3^3\times13}{-13}\)

                                      \(=-27\)

Cảm Ơn !

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}=2\)

\(\Rightarrow1+\frac{1}{2.3}.2+\frac{1}{3.4}.2+...+\frac{1}{x\left(x+1\right)}.2=2\)

=> \(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x\left(x+1\right)}=1\)

=> \(1-\frac{1}{x+1}=1\)

=> \(\frac{1}{x+1}=0\Rightarrow0\left(x+1\right)=1\Rightarrow x\in\varnothing\)

\(\frac{1}{1.2:2}+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{x.\left(x+1\right):2}=2\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}=2\)

\(2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2\)

\(1-\frac{1}{x+1}=1\)

\(\frac{1}{x+1}=0\)

Vậy x vô nghiệm.

\(\frac{30303}{80808}=\frac{3\times10101}{8\times10101}=\frac{3}{8}\)

\(3^{x+4}+3^{x+2}=270\)

\(< =>3^x.81+3^x.9=270\)

\(< =>3^x=\frac{270}{90}=3< =>x=1\)

\(3^{x+4}+3^{x+2}=270\)

\(\Leftrightarrow3^x.81+3^x.9=270\)

\(\Leftrightarrow3^x\left(81+9\right)=270\)

\(\Leftrightarrow3^x.90=270\)

\(\Leftrightarrow3^x=3\)

\(\Leftrightarrow x=1\)

Đặt A=2.2².2³.2⁴...2¹⁰⁰

2A=2(2.2².2³.2⁴...2¹⁰⁰)

2A=2².2³.2⁴...2¹⁰¹

2A-A=2¹⁰¹-2

A=2¹⁰¹-2/2

Động Từ :

Put , do , pick , go , get , have , study , sing , catch , dance

Danh từ :

dog , cat , hat , shirt , balloon , print , piano , bass , computer , maracas

10 động từ là:go,jum,sing,dance,sleep,water,wash,write,read,cry

9 danh từ là:flower,sand,family,home,house,book,pen,computer,clock

ks nhé!Học tốt!

\(I=1^2+2^2+3^2+4^2+...+2017^2+2018^2\)

\(I=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)

\(I=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)

\(I=1+2+3+4+...+2017+2018\)

\(I=\frac{\left(2018+1\right).2018}{2}=2037171\)

I = 1² + 2² + 3² + ... + 2017²

= 1.1 + 2.2 + 3.3 + ... + 2017.2017

= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 2017.(2018 - 1)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - (1 + 2 + 3 + 4 + ... + 2017)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - 2035153

Đặt K = 1.2 + 2.3 + 3.4 + ... + 2017.2018

=> 3K = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2017.2018.3

=> 3K = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.2018.(2019 - 2016)

=> 3K = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2017.2018.2019 - 2016.2017.2018

=> 3K = 2017.2018.2019

=> K = 2017.2018.2019 : 3 

=> K = 2739315938

Lại có I = K - 2035153

= 2739315938 - 2035153

= 2 737 280 785

M = \(54-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{12}\left(1+2+3+...+12\right)\)

\(54-\left(\frac{3}{2}+\frac{4}{2}+...+\frac{13}{2}\right)=54-\frac{1}{2}\left(3+4+5+...+13\right)=54-\frac{1}{2}.176=54-88\)

= -24

\(I=1\left(2-1\right)+2\left(3-1\right)+...+2017\left(2018-1\right)\)

\(I=\left(1.2+2.3+...+2017.2018\right)-\left(1+2+...+2017\right)\)

\(3I=\left(1.2.3+2.3\left(4-1\right)+...+2017.2018\left(2019-2016\right)\right)-\frac{3.2017.2018}{2}\)

=> \(I=\frac{2017.2018.2019}{3}-2017.1009\)

=> \(I=.....\)