Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left[3^{22}. 3^5+2.3^{27}\right]:3^{26}\)
\(=\left[3^{27}+2.3^{27}\right]:3^{26}\)
\(=\left[3^{27}.\left(1+2\right)\right]:3^{26}\)
\(=\left[3^{27}.3\right]:3^{26}\)
\(=3^{28}:3^{26}\)
\(=3^2\)
\(\left[3^{22}\cdot\left(3^8:3^3\right)+2\cdot3^{27}\right]:3^{26}\\ =\left(3^{22}\cdot3^5+2\cdot3^{27}\right):3^{26}\\ =3^{27}\left(1+2\right):3^{26}\\ =3^{27}\cdot3:3^{26}\\ =3^2=9\)
\(\left[3^{22}.\left(3^8:3^3\right)+2.3^{27}\right]\):\(3^{26}\)
=\(\left(3^{22}.3^5+2.3^{27}\right):3^{26}\)
=(\(3^{27}\)+2.\(3^{27}\)):\(3^{26}\)
=\(3^{27}\).(1+2):\(3^{26}\)
=\(3^{28}:3^{26}=3^2\)=9
a. (7-3+1)/8 = 5/8
b. (-5+26-1)/21 = 20/21
c. (-3-8+5)/27 = -6/27= -2/9
\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)
cho A=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
=> n-A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
=>\(3\left(n-A\right)\)=\(1\)\(+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{3n-1}}\)
=> \(3\left(n-A\right)-\left(n-A\right)=2\left(n-A\right)=1-\frac{1}{3^n}\)
=>\(2\left(n-A\right)< 1\)
=>\(n-A< \frac{1}{2}\)
=> \(A< n-\frac{1}{2}\)
Deu la tui het do
Lời giải:
\(P=1+2+22+23+24+25+26+27\)
\(=(22+23)+24+(25+2)+(26+1)+27\)
\(=45+24+27+27+27=3.15+3.8+3.27\)
\(=3(15+8+27)\vdots 3\)