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K/c mỗi số là 3 đơn vị.
Tổng đó có số số hạng là:`(40-1):3+1=14`(số)
`=>1+4+7+...+40=((40+1).14)/2=41.7=287`
\(A=\)\(1+4+7+10+...+34+37+40\)
\(A=\)\(40+37+34+31+...+7+4+1\)
\(2A=\)\(41+41+41+41+...+41+41+\text{41}\)
\(2A=41.14=>A=\dfrac{41.14}{2}=287\)
\(A=\dfrac{30}{4x7}+\dfrac{30}{7x10}+\dfrac{30}{10x13}+...+\dfrac{30}{37x40}\)
\(=10x\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+10x\left(\dfrac{1}{7}-\dfrac{1}{10}\right)+...+10x\left(\dfrac{1}{37}-\dfrac{1}{40}\right)\)
=10x\(\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{10x9}{10x4}=\dfrac{9}{4}\)
Nguyễn Quốc Thịnhanswer : 287
integrater his light Nguyễn Quốc Thịnh
Mình nói Tiếng Anh
1.\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{23}-\frac{4}{27}\)
\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)
2. Đặt \(A=\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+\frac{3}{374}+\frac{3}{594}+\frac{3}{864}\)
\(\Rightarrow A=\frac{3}{2.7}+\frac{3}{7.12}+...+\frac{3}{27.32}\)
\(\Rightarrow5A=3.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{27.32}\right)\)
\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)\)
\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(\Rightarrow5A=3.\frac{15}{32}=\frac{45}{32}\Rightarrow A=\frac{45}{32}:5=\frac{9}{32}\)
3. Đặt \(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\)
\(\Rightarrow3S=\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)
\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)
Câu 1:
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 37 + 38 - 39 - 40
= ( 1+2-3-4 ) + ( 5+6-7-8 ) + ... + ( 37+38-39-40)
= (-4) + (-4) +... + (-4)
= (-4) . 10 (DO CÓ 10 NHÓM)
= -40
a. 37-7.(x+1)=40-8.3
37 - 7(x+1) = 16
7(x+1) = 21
x+1 = 3
x=2
b. (x+5).3=11=10+(3+4).4
(sao có 2 dấu bằng, dấu nào là dấu cộng do viết nhầm ???)
c. 25-5.3+4.10=100-5.x
50 = 100 - 5x
5x = 50
x=10
d. 36+2.(x-7)=12+8.(3+5)-36
36 + 2(x-7) = 40
2(x-7) = 4
x-7 =2
x=9
e. 3.(x+7)-9=11.5-5-8
3(x+7) - 9 = 42
3(x+7) = 51
x+7 = 17
x=10
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
Yêu cầu là gì vậy?????
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