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a/
$\frac{7}{4}+\frac{9}{16}=\frac{28}{16}+\frac{9}{16}=\frac{28+9}{16}=\frac{37}{16}$
b/
$\frac{-70}{27}+\frac{35}{9}=\frac{-70}{27}+\frac{105}{27}=\frac{35}{27}$
c/
$\frac{-4}{-5}+\frac{-2}{3}=\frac{4}{5}-\frac{2}{3}=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}$
d/ Đề thiếu. Bạn xem lại.
i/
$\frac{-7}{2}+(\frac{-5}{-4}-\frac{4}{-9}$
$=\frac{-7}{2}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-14}{4}+\frac{5}{4}+\frac{4}{9}$
$=\frac{-9}{4}+\frac{4}{9}=\frac{-81}{36}+\frac{16}{36}=\frac{-65}{36}$
k/
$\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$=\frac{3}{12}+\frac{4}{12}-\frac{6}{12}=\frac{1}{12}$
m/
$\frac{-1}{9}-\frac{1}{7}-\frac{3}{4}$
$=-(\frac{1}{9}+\frac{1}{7}+\frac{3}{4})$
$=-(\frac{28}{252}+\frac{36}{252}+\frac{189}{252})$
$=\frac{-253}{252}$
n/
$\frac{11}{12}-(\frac{-3}{18}+\frac{5}{6}$
$=\frac{11}{12}+\frac{1}{6}+\frac{5}{6}$
$=\frac{11}{12}+1=\frac{23}{12}$
o/
$\frac{20}{17}+\frac{-11}{2}-\frac{30}{4}$
$=\frac{20}{17}-\frac{11}{2}-\frac{15}{2}$
$=\frac{20}{7}-13=\frac{-71}{7}$
v/
$\frac{1}{2}-\frac{3}{4}+\frac{5}{6}$
$=\frac{6}{12}-\frac{9}{12}+\frac{10}{12}$
$=\frac{6-9+10}{12}=\frac{7}{12}$
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Lời giải:
$A=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....+\frac{69}{7^{70}}$
$7A=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$
$\Rightarrow 6A=7A-A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{69}}-\frac{69}{7^{70}}$
$42A=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$
$\Rightarrow 36A=42A-6A=1-\frac{69}{7^{69}}+\frac{69}{7^{70}}<1$
$\Rightarrow A< \frac{1}{36}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
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bài 1:
gọi số lớn là a , số bé là b
Theo đề cho , ta có :
a - b = 33 (1 )
\(\frac{a}{b}\) = 3 => a=3b (2)
Thay (2) vào (1) ta có :
a - b = 33 ↔ 3b - b = 33 ↔ 2b = 33 → b = 33 : 2 = 16,5
a = 3b → a = 3.16,5 = 49,5
Vậy số lớn là 49,5
số bé là 16,5
bài 2 :
a) 58 . 75 + 58 . 50 - 58 . 25
= 58 . ( 75 + 50 - 25 )
= 58 . 100 = 5800
b) 27 . 121 - 87 . 27 + 73 . 34
= 27 . ( 121 - 87 ) + 73 . 34
= 27 . 34 + 73 . 34
= 34 . ( 27 + 73 )
= 34. 100 = 3400
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\(A=1\cdot2+2\cdot3+...+68\cdot69+69\cdot70\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+69\cdot70\cdot\left(71-68\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+69\cdot70\cdot71-68\cdot69\cdot70\)
\(3A=69\cdot70\cdot71\Rightarrow A=\dfrac{69\cdot70\cdot71}{3}=114310\)
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2x - 138 = 23 . 32
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 105
10 + 2x = 45 : 43
10 + 2x = 16
2x = 16 - 10
2x = 6
x= 3
a) \(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\).
b) Áp dụng kết quả câu a) ta được:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{69.70}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{69}-\frac{1}{70}\)
\(=1-\frac{1}{70}\)
\(=\frac{69}{70}\).