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a: \(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}=\dfrac{15}{26}-\dfrac{4}{26}=\dfrac{11}{26}\)
b: \(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-5=\dfrac{-4}{7}-5+\dfrac{2}{9}=-\dfrac{337}{63}\)
c: \(=-\dfrac{11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
\(a)\) Công thức tính số hạng của một dãy số là : (Số cuối-số đầu ) chia khoảng cách rồi cộng thêm 1 .
Do đó : Số hạng của dãy số A là : \(\dfrac{\left(2n+1\right)-1}{2}+1=n+1\)
Số hạng của dãy số B là : \(\dfrac{2n-2}{2}+1=n-1+1=n\)
\(b)\) Ta có : Số hạng của dãy số A là : \(n+1\)
Do đó : tổng của A là : \(\dfrac{\left(2n+1+1\right).\left(n+1\right)}{2}=\dfrac{2\left(n+1\right)\left(n+1\right)}{2}\)
\(=\left(n+1\right)^2\)
Vì n thuộc N nên tổng của A là : một số chính phương .
\(c)\) Ta có : Số hạng của dãy số B là : n
Do đó : Tổng của dãy số B là : \(\dfrac{n.\left(2n+2\right)}{2}=\dfrac{2.n.\left(n+1\right)}{2}\)
\(=n.\left(n+1\right)\)
Ta thấy : n(n+1) là tích của 2 số tự nhiên liên tiếp nên để B là số chính phương thì khi và chỉ khi n hoặc n+1 bằng 0 .
Ta thấy chúng đều không thoả mãn .
vậy.............
Giải:
a) \(\left(-\dfrac{5}{28}+1,75+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(-\dfrac{5}{28}+\dfrac{7}{4}+\dfrac{8}{35}\right):\left(\dfrac{-39}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9}{5}:\left(-\dfrac{39}{20}\right)\)
\(=\dfrac{9.\left(-20\right)}{5.39}\)
\(=\dfrac{3.\left(-4\right)}{1.13}\)
\(=\dfrac{-12}{13}\)
b) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{22}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
c) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-528.\dfrac{2}{15}\)
\(=70,5-\dfrac{1056}{15}\)
\(=70,5-70,4\)
\(=0,1\)
a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)
\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)
\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)
\(x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\).
b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)
\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)
\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)
\(3x-3\cdot7=-\dfrac{19}{2}\)
\(3x-21=-\dfrac{19}{2}\)
\(3x=-\dfrac{19}{2}+21\)
\(3x=\dfrac{23}{2}\)
\(x=\)\(\dfrac{23}{2}:3\)
\(x=\dfrac{23}{6}\)
Vậy \(x=\dfrac{23}{6}\).
c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)
\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)
\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)
\(\dfrac{3}{4x}=-\dfrac{65}{22}\)
\(4x=\dfrac{3\cdot22}{-65}\)
\(4x=-\dfrac{66}{65}\)
\(x=-\dfrac{66}{65}:4\)
\(x=-\dfrac{33}{130}\)
Vậy \(x=-\dfrac{33}{130}\).
d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(-\dfrac{2}{3}x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:-\dfrac{2}{3}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\).
e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)
\(\left|x\right|=\dfrac{29}{12}\)
\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)
Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
a, \(\left(4\dfrac{1}{9}+3\dfrac{1}{4}\right).2\dfrac{1}{4}+2\dfrac{3}{4}\)
\(=\left(\dfrac{37}{9}+\dfrac{13}{4}\right).\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265}{36}.\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\dfrac{265}{16}+\dfrac{11}{4}\)
\(=\dfrac{309}{16}\)
b, \(\dfrac{9}{23}.\dfrac{5}{8}+\dfrac{9}{23}.\dfrac{3}{8}-\dfrac{9}{23}\)
\(=\dfrac{45}{184}+\dfrac{27}{184}-\dfrac{9}{23}\)
\(=\dfrac{9}{23}-\dfrac{9}{23}\)
\(=\dfrac{1}{1}\)
c, \(1+\left(\dfrac{9}{10}-\dfrac{4}{5}\right)\div3\dfrac{1}{6}\)
\(=1+\left(\dfrac{9}{10}-\dfrac{4}{5}\right)\div\dfrac{19}{6}\)
\(=1+\dfrac{1}{10}\div\dfrac{19}{6}\)
\(=1+\dfrac{3}{95}\)
\(=1\dfrac{3}{95}\)
d, ???
a)\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)=\(7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)=\(\left(7\frac{4}{9}-3\frac{4}{9}\right)+4\frac{7}{11}\)= 4+\(4\frac{7}{11}\)=\(8\frac{7}{11}\)
b)\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)=\(\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5+\frac{7}{9}\)=\(\frac{-7}{9}.1+5+\frac{7}{9}\)=\(\frac{-7}{9}+\frac{7}{9}+5\)=\(\left(\frac{-7}{9}+\frac{7}{9}\right)+5\)= 0+5=5
c)\(50\%.1\frac{1}{3}.10\frac{7}{35}.0,75\)= \(\frac{1}{2}.\frac{4}{3}.10\frac{1}{5}.\frac{3}{4}\)=\(\frac{1}{2}.\frac{4}{3}.\frac{51}{5}.\frac{3}{4}\)=\(\frac{1.4.51.3}{2.3.5.4}\)=\(\frac{51}{2.5}\)=\(\frac{51}{10}\)
d)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)=\(\frac{43}{43}-\frac{1}{43}\)=\(\frac{42}{43}\)
a) \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=\left(\frac{67}{9}+\frac{51}{11}\right)-\frac{31}{9}\)
\(=\left(\frac{67}{9}-\frac{31}{9}\right)+\frac{51}{11}\)
\(=\frac{36}{9}+\frac{51}{11}\)
\(=\frac{95}{11}=8\frac{7}{11}\)
b) \(-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(=-\frac{7}{9}.\frac{4}{11}+-\frac{7}{9}.\frac{7}{11}+\frac{52}{9}\)
\(=-\frac{7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)
\(=-\frac{7}{9}.1+\frac{52}{9}\)
\(=-\frac{7}{9}+\frac{52}{9}\)
= 5
d) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=1.\left(1-\frac{1}{43}\right)\)
\(=1.\frac{42}{43}\)
\(=\frac{42}{43}\)
a) Ta có:
\(10^{10}=10...0\Rightarrow10^{10}-1=10..0-1=9..99\)
Nên \(10^{10}-1\) ⋮ 9
b) Ta có:
\(10^{10}=10...0\Rightarrow10^{10}+2=10..0+2=10..2\)
Mà: \(1+0+0+...+2=3\) ⋮ 3
Nên: \(10^{10}+2\) ⋮ 3
