Đề bài yêu cầu gì hả bạn.

tính kết qủa á

S = 2²⁰¹⁶ + 2²⁰¹³ + 2²⁰¹⁰ + ... + 2³ + 1

8S = 2²⁰¹⁹ + 2²⁰¹⁶ + 2²⁰¹³ + ... + 2⁶ + 2³

⇒ 7S = 8S - S

= (2²⁰¹⁹ + 2²⁰¹⁶ + 2²⁰¹³ + ... + 2⁶ + 2³) - (2²⁰¹⁶ + 2²⁰¹³ + 2²⁰¹⁰ + ... + 2³ + 1)

= 2²⁰¹⁹ - 1

⇒ S = (2²⁰¹⁹ - 1) : 7

Hai điểm cùng phía với M là: 0; N

   -49 + 118 - 52

= -(49 + 52) + 118

= -101 + 118

= 17

Ta có:

\(\dfrac{1}{430}\)+\(\dfrac{1}{324}\)

=\(\dfrac{162}{69660}\)+\(\dfrac{215}{69660}\)

=\(\dfrac{162+215}{69660}\)

=\(\dfrac{377}{69660}\)

\(\dfrac{1}{430}+\dfrac{1}{324}\)

\(=\dfrac{162}{69660}+\dfrac{215}{69660}\)

\(=\dfrac{377}{69660}\)

\(\dfrac{1}{555}+\dfrac{1}{678}\)

\(=\dfrac{678}{376290}+\dfrac{555}{376290}\)

\(=\dfrac{1233}{376290}=\dfrac{411}{125430}=\dfrac{137}{41810}\)

Ta có:

\(\dfrac{1}{555}\)+\(\dfrac{1}{678}\)

=\(\dfrac{678}{376290}\)+\(\dfrac{555}{376290}\)

=\(\dfrac{678+555}{376290}\)

=\(\dfrac{1233}{376290}\)

=\(\dfrac{137}{41810}\)

Ta thấy: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)

\(\dots\)

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

Suy ra: \(A=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots+\dfrac{1}{100^2}\)

\(< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dots+\dfrac{1}{99\cdot100}\)

\(=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=2-\dfrac{1}{100}< 2\)

Vậy \(A< 2\)

Số số hạng của tổng:

(584 - 3) : 7 + 1 = 84 (số)

3 + 10 + 17 + ... + 584 = (584 + 3) . 84 : 2 = 24654

A = 3 + 10 + 17 +...+ 584

Dãy số trên là dãy số cách đều với khoảng cách là: 10 -  3  = 7

Số số hạng của dãy số trên là: (584 - 3) : 7 + 1  = 84

Tổng của dãy số trên là:

A = (584 + 3)  x 84 : 2 = 24654

Vậy 3 + `10 + 17 +...+ 584 = 24654

a, \(\dfrac{42}{54}=\dfrac{7}{x}\)

Ta có: \(x.42=7.54\)

\(=>x.42=378\)

\(=>x=378:42\)

\(=>x=9\)

Vậy x = 9

b, \(\dfrac{-2}{3}=\dfrac{y}{15}\)

Ta có: \(y.3=\left(-2\right).15\)

\(=>y.3=-30\)

\(=>y=\left(-30\right):3\)

\(=>y=-10\)

Vậy y = -10

c, \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)

* Ta có: \(x.6=3.10\)

\(=>x.6=30\)

\(=>x=30:6\)

\(=>x=5\)

Vì  x = 5 \(\Rightarrow\dfrac{3}{5}=\dfrac{y}{-20}\)

Ta có: \(y.5=3.\left(-20\right)\)

\(=>y.5=-60\)

\(=>y=\left(-60\right):5\)

\(=>y=-12\)

Vậy x = 5 ; y = -12

d, \(\dfrac{-x}{-6}=\dfrac{-5}{6}\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\Rightarrow x=-5\) ( Cùng mẫu số )

Vậy x = -5

\(#NqHahh\)

\(a.\) \(\dfrac{42}{54}=\dfrac{7}{x}\)

\(\Rightarrow x\cdot42=7\cdot54\)

\(\Rightarrow x\cdot42=378\)

\(\Rightarrow x=378:42\)

\(\Rightarrow x=9\)

Vậy \(\dfrac{42}{54}=\dfrac{7}{9}.\)

\(b.\) \(\dfrac{-2}{3}=\dfrac{y}{15}\)

\(\Rightarrow y\cdot3=\left(-2\right)\cdot15\)

\(\Rightarrow y\cdot3=\left(-30\right)\)

\(\Rightarrow y=\left(-30\right):3\)

\(\Rightarrow y=\left(-10\right)\)

Vậy \(\dfrac{-2}{3}=\dfrac{-10}{15}\)

\(c.\) \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)

\(\Rightarrow x\cdot6=3\cdot10\)

\(\Rightarrow x\cdot6=30\)

\(\Rightarrow x=30:6\)

\(\Rightarrow x=5\)

Vậy: \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{y}{-20}\)

Mặt khác: \(\dfrac{3}{5}=\dfrac{y}{-20}\)

\(\Rightarrow y\cdot5=3\cdot\left(-20\right)\)

\(\Rightarrow y\cdot5=\left(-60\right)\)

\(\Rightarrow y=\left(-60\right):5\)

\(\Rightarrow y=\left(-12\right)\)

Vậy \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{-12}{-20}\)

\(d.\) \(\dfrac{-x}{-6}=\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\)

Do cùng mẫu số nên ta xét tử, ta thấy:

\(x=\left(-5\right)\)

Vậy \(\dfrac{-5}{6}=\dfrac{-5}{6}\)

Ta có: \(8b-9a=31\)

\(\Rightarrow8b=31+9a\)

\(\Rightarrow b=\dfrac{31+9a}{8}\)

\(\Rightarrow b=\dfrac{32+8a+a-1}{8}\)

\(\Rightarrow b=\dfrac{8\cdot\left(4+a\right)+a-1}{8}\)

\(\Rightarrow b=4+a+\dfrac{a-1}{8}\)

Để \(b\in N\) thì:

\(\dfrac{a-1}{8}\in N\)

\(\Rightarrow a-1⋮8\)

\(\Rightarrow a-1=8k\left(k\in N\right)\)

\(\Rightarrow a=8k+1\)

Khi đó: \(b=4+8k+1+\dfrac{8k+1-1}{8}\)

\(\Rightarrow b=5+8k+\dfrac{8k}{8}\)

\(\Rightarrow b=5+8k+k\)

\(\Rightarrow b=5+9k\)

Mặt khác: \(\dfrac{11}{17}< \dfrac{a}{b}< \dfrac{23}{29}\)

\(\Rightarrow\dfrac{11}{17}< \dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)

Xét: \(\dfrac{11}{17}< \dfrac{8k+1}{5+9k}\)

\(\Rightarrow11\left(5+9k\right)< 17\left(8k+1\right)\)

\(\Rightarrow55+99k< 136k+17\)

\(\Rightarrow136k-99k>55-17\)

\(\Rightarrow37k>38\)

\(\Rightarrow k>\dfrac{38}{37}\left(1\right)\)

Xét: \(\dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)

\(\Rightarrow29\left(8k+1\right)< 23\left(5+9k\right)\)

\(\Rightarrow232k+29< 115+207k\)

\(\Rightarrow232k-207k< 115-29\)

\(\Rightarrow25k< 86\)

\(\Rightarrow k< \dfrac{86}{25}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{38}{27}< k< \dfrac{86}{25}\)

Mà \(k\in N\)

\(\Rightarrow k\in\left\{2;3\right\}\)

\(+,\) \(k=2\).

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot2+1=17\\b=5+9\cdot2=23\end{matrix}\right.\)

\(+,\) \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot3+1=25\\b=5+9\cdot3=32\end{matrix}\right.\)

\(\Rightarrow\) Vậy \(\left(a;b\right)\in\left\{\left(17;23\right),\left(25;32\right)\right\}\)