xam xi

\(B=2^{2018}-2^{2017}-2^{2016}-2^{2015}-2^{2014}\)

\(=>2B=2^{2019}-2^{2018}-2^{2017}-2^{2016}-2^{2015}\)

\(=>2B+B=2^{2019}-2^{2014}\)

\(=>B=\dfrac{2^{2019}-2^{2014}}{3}\)

\(S=1-2+2^2-2^3+...+2^{2012}-2^{2013}\)

\(\Rightarrow2S=2-2^2+2^3-2^4+...+2^{2013}-2^{2014}\)

\(\Rightarrow2S+S=2-2^2+2^3-...-2^{2014}+1-2^2-2^3+...-2^{2013}\)

\(\Rightarrow3S=1-2^{2014}\)\(\Rightarrow3S-2^{2014}=1-2^{2015}\)

Hình như sai đề bạn ạ!! Thiếu 3² thì phải.

Đặt \(A=2^{2013}+2^{2012}+\cdots+2^3+2^2+3\)

=>\(A=2^{2013}+2^{2012}+\cdots+2^2+2+1\)

=>\(2A=2^{2014}+2^{2013}+\cdots+2^3+2^2+2\)

=>\(2A-A=2^{2014}+2^{2013}+\cdots+2^3+2^2+2-2^{2013}-2^{2012}-\cdots-2^2-2-1\)

=>\(A=2^{2014}-1\)

Ta có: \(B=2^{2014}-2^{2013}-2^{2012}-\cdots-2^3-2^2-3\)

=>\(B=2^{2014}-\left(2^{2014}-1\right)\)

\(=2^{2014}-2^{2014}+1=1\)

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

a: \(3\frac{4}{27}+\frac{11}{21}+0,5-\frac{4}{27}+\frac{10}{21}\)

\(=\left(3+\frac{4}{27}-\frac{4}{27}\right)+0,5+\left(\frac{11}{21}+\frac{10}{11}\right)\)

=3+0,5+1

=4,5

b: \(6-3\cdot\left(-\frac13\right)^3-\sqrt9\)

\(=6-3\cdot\frac{-1}{27}-3\)

\(=3+\frac19=\frac{27}{9}+\frac19=\frac{28}{9}\)

c: \(\left(5,3-2,8\right)-\left(4+5,3\right)+0,2\)

=5,3-2,8-4-5,3+0,2

=-2,8-4+0,2

=-6,8+0,2

=-6,6

d: \(0,5\cdot\sqrt{100}-\sqrt{\frac14}=0,5\cdot10-\frac12=5-0,5=4,5\)

e: \(13\frac12\cdot\frac57-\frac57\cdot27\frac12\)

\(=\frac57\left(13+\frac12-27-\frac12\right)\)

\(=\frac57\cdot\left(-14\right)=-10\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27-28-\frac27\right):\left(-\frac35\right)\)

\(=-12\cdot\left(-\frac53\right)=20\)

g: \(\sqrt{0,01}-\sqrt{0,25}\)

=0,1-0,5

=-0,4