P lớn hơn Q nha

a)\(A=2+2^2+2^3+...+2^{99}+2^{100}\Leftrightarrow\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

\(\Leftrightarrow6+2^2\left(2+2^2\right)+2^4\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)

\(\Leftrightarrow6+2^2\cdot6+2^4\cdot6+...+2^{98}\cdot6\)

\(\Leftrightarrow6\left(1+2^2+2^4+...+2^{98}\right)⋮6\)

Mà tổng đó chia hết cho 6 thì cũng chia hết cho 2 và 3

b) \(A=2+2^2+2^3+...+2^{99}+2^{100}\)

\(\Leftrightarrow\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(\Leftrightarrow\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow\left(2+2^2+2^3+2^4\right)\left(1+2^4+...+2^{96}\right)\)

\(\Leftrightarrow6\cdot5\left(1+2^4+...+2^{96}\right)⋮5\)

Mà tổng này chia hết cho cả 2 và 5 hay tổng đó chia hết cho 10 mà số nào chia hết cho 10 cũng tận cùng là chữ số 0

\(\Rightarrow\)Chữ số tận cùng của A là 0

101.(-162)+38(-101)+101

= 101.(-162)-38.101+101=101.(-162-38+1)=101.(-199)=-20099

ko ai rả lời thì thôi vậy

Ta có : A=\(\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)

= \(2.\left(\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{48.53}\right)\)

= \(\frac{2}{5}.\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)

=\(\frac{2}{5}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{48}-\frac{1}{53}\right)\)

=\(\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)

=\(\frac{2}{5}.\frac{50}{159}\)

=\(\frac{20}{159}\)

Vậy A=\(\frac{20}{159}\)

\(A=\frac{2}{3.8}+\frac{2}{8.13}+...+\frac{2}{48.53}\)

\(=\frac{2}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{48.53}\right)\)

\(=\frac{2}{5}\left(\frac{8-3}{3.8}+\frac{13-8}{8.13}+...+\frac{53-48}{48.53}\right)\)

\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{48}-\frac{1}{53}\right)\)

\(=\frac{2}{5}\left(\frac{1}{3}-\frac{1}{53}\right)\)

\(=\frac{20}{159}\)

\(3-\left(17-x\right)=2819-\left(36+289\right)\)

\(3-\left(17-x\right)=2819-\left(36+289\right)\)

\(17-x=3-2494\)

\(17-x=-2491\)

\(x=-2491-17\)

\(x=-2508\)

Câu đầu tiên mẫu số \(9\)mũ là gì vậy

Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)

Ta lại có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)

Từ (1) và (2) suy ra đpcm.

\(\frac{-8}{9}+\frac{1}{9}\times\frac{2}{9}+\frac{1}{9}\div\frac{7}{9}.\)

\(\Rightarrow\left(\frac{1}{9}+\frac{1}{9}\times\frac{2}{9}\right)+\frac{-8}{9}\div\frac{7}{9}\)

\(\Rightarrow1+\frac{-8}{7}\)

\(\Rightarrow\frac{-5}{7}\)