a = 1/3+1/6+1/10+1/15 ......+1/45
ghi lời giải rõ ràng
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NV
10 tháng 7 2018
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}\)\(+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(=\frac{1}{4}+\frac{1}{3}\)
\(=\frac{7}{12}\)
12 tháng 9 2016
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
\(=>A=\frac{1\cdot2+4\cdot1\cdot2+9\cdot1\cdot2+16\cdot1\cdot2+25\cdot1\cdot2}{3\cdot4+4\cdot3\cdot4+9\cdot3\cdot4+16\cdot3\cdot4+25\cdot3\cdot4}\)
\(=>A=\frac{\left(1+4+9+16+25\right)\cdot1\cdot2}{\left(1+4+9+16+25\right)\cdot3\cdot4}=\frac{1}{6}=\frac{111111}{666666}\)
Mà \(\frac{111111}{666666}< \frac{111111}{666665}\)
\(=>A< B\)
H
14 tháng 9 2021
\(6+\frac{4}{12}\)\(-\frac{2}{15}\)
\(C1:6+\left(\frac{1}{3}-\frac{2}{15}\right)\)
\(\frac{90}{15}+\left(\frac{5}{15}-\frac{2}{15}\right)\)\(=\frac{93}{15}\)\(=\frac{31}{5}\)
\(C2:6+\frac{4}{12}-\frac{2}{15}\)
\(6+\frac{1}{3}+\frac{2}{15}\)
\(\frac{90}{15}+\frac{5}{15}-\frac{2}{15}\)
\(b\)Tương tự như câu a em tự giải coi đó như bài tập nha
Phá ngoặc trc nó là dấu cộng thì giữa nguyên dấu
trc nó là dấu trừ thì đổi dấu các số hạng trong ngoặc
\(\)
14 tháng 9 2021
\(6+\left(\frac{4}{12}-\frac{2}{15}\right)=6+\left(\frac{5}{15}-\frac{2}{15}\right)=\frac{30}{5}+\frac{1}{5}=\frac{31}{5}\)
\(1+\left(\frac{5}{6}+\frac{7}{3}+\frac{2}{12}\right)=1+\left(\frac{10+28+2}{12}\right)=1+\frac{40}{12}=\frac{3}{3}+\frac{10}{3}=\frac{13}{3}\)
20 tháng 2 2017
Ta có : \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
12 tháng 1 2016
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=>2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\)
A= 1/2(1/3 - 1/101)
A= 49/303
12 tháng 1 2017
Bài 1:
\(a,\left(a+b-c\right)-\left(b-c-d\right)\)
\(=a+b-c-b+c+d\)
\(=a+d\)
\(b,-\left(a-b+c\right)+\left(a-b+d\right)\)
\(=-a+b-c+a-b+d\)
\(=-c+d\)
\(c,\left(a+b\right)-\left(-a+b-c\right)\)
\(=a+b+a-b+c\)
\(=2a+c\)
\(d,-\left(a+b\right)+\left(a+b+c\right)\)
\(=-a-b+a+b+c\)
\(=c\)
Bài 3 :
\(a,15-\left(4-x\right)=6\)
\(4-x=15-6\)
\(4-x=9\)
\(x=4-9\)
\(x=-5\)
\(b,-30+\left(25-x\right)=-1\)
\(25-x=-1+30\)
\(25-x=29\)
\(x=25-29\)
\(x=-4\)
\(c,x-5=-1\)
\(x=-1+5\)
\(x=4\)
\(d,x-4=-10\)
\(x=-10+4\)
\(x=-6\)
\(e,x+3=-8\)
\(x=-8-3\)
\(x=-11\)
\(g,x+6=0\)
\(x=-6\)
12 tháng 1 2017
Câu 1:
A, (a+b-c)-(b-c-d)
= a+b-c-b+c+d
= a+(b-b)+(c-c)
= a
B, -(a-b+c)+(a-b+d)
= -a+b-c+a+b+d
= (a-a)+(b+b)+d-c
= 2b+d-c
C, (a+b)-(-a+b-c)
= a+b+a-b+c
= (a+a)+(b-b)+c
= 2a+c
D, -(a+b)+(a+b+c)
= -a-b+a+b+c
= (-a+a)+(b-b)+c
= c
=
5 tháng 8 2023
\(1-5+9-13+17-...+201-205\\ =\left(1+9+17+201\right)-\left(5+13+...+205\right)\\ =M-N\left(1\right)\)
Xét \(M=1+9+17+...+201\left(26sh\right)\\ =\dfrac{\left(201+1\right)\cdot26}{2}=2626\left(2\right)\)
Xét \(N=5+13+...+205\left(26sh\right)\\ =\dfrac{\left(205+5\right)\cdot26}{2}=2730\left(3\right)\)
Từ (1);(2) và (3)\(\Rightarrow2626-2730=-104\)
\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)
\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
=> \(a=2.\frac{2}{5}\)
=> \(a=\frac{4}{5}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)