a) Chứng minh: \(\frac{1}{a\left(a+1\right)}=\frac{1}{a}+\frac{-1}{a+1}\)(a ∈ N*)
b) Tính: B =\(\frac{11}{5.7}+\frac{11}{7.9}+\frac{11}{9.11}+...+\frac{11}{59.61}\)
Các bạn giúp mình với! Mai mình phải nộp bài rồi:(((
Tính hợp lí:
\(a)\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(b)1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-.....-\frac{2}{97.99}\)
Giúp mình với! Mình cần gấp!
Ai nhanh mình tích cho!
a, Ta có:
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)
k đúng cho mình nha. Thanks!!!
a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.
\(=\frac{2}{3}\times\frac{0,1-\frac{1}{9}+\frac{1}{11}}{0,1-\frac{1}{9}+\frac{1}{11}}+\frac{-\frac{2}{3}\times\left(-1-\frac{3}{7}+\frac{3}{28}\right)}{-1-\frac{3}{7}+\frac{3}{28}}\)
=\(\frac{2}{3}+\left(-\frac{3}{2}\right)\)
=\(-\frac{5}{6}\)
tìm số nguyên x
a) (x-3)+(x-2)+(x-1)+....+10+11=11
b) \(\frac{2}{3}\).\(\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]\)<= x <=\(4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
giúp mình nha mai mình đi học rồi. minh cần gấp lắm bạn nào giải được thì mau giải giúp mình nha
Bài 1:Tính bằng cách hợp lý:
a)\(\frac{8.4.125.25+96524+3476}{1+5+9+...+61+62,5-26}\)
b)\(\frac{5+55+555+5555}{9+99+999+9999}\)
c)39,2.27+39,2.43+78,4.15
d)\(\frac{4}{17}.\frac{3}{11}+\frac{8}{11}.\frac{4}{17}-\frac{4}{17}\)
e)(812,1.10,5-91.10,5).(41.11-4100.0,1-41)
Bài 2:Tính nhanh các tổng sau:
a)\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{57.59}\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2012}\right)\)
GẤP!GẤP!GẤP!MAI MIK NỘP RỒI!BẠN NÀO ĐẦU TIÊN MIK TIK CHO NHA!
*Lưu ý:Dấu chấm là dấu nhân đó
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
Thực hiện phép tính :
a, \(\sqrt{12\frac{1}{4}}.\left(-\frac{2}{7}\right)^2-\left[2,\left(4\right).2\frac{5}{11}\right]:(-\frac{42}{5})\)
b, \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2007}\right)\)
c, \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
Các bạn giải nhanh giúp mình nha!
tính một cách hợp lí
a.\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
lưu ý phép tính a ko có phép nhân VD' 4\(\frac{3}{4}\) LÀ hỗn số nhé
b.\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
c.\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
a.\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(2,5\right)+3\frac{1}{12}\)
\(=\)\(\left(4\frac{3}{4}+\frac{1}{8}+3\frac{1}{12}\right)-\left(0,37+1,28+2,5\right)\)
\(=7\frac{23}{24}-4,15\)
\(=7\frac{23}{24}-4\frac{3}{20}\)
\(=3\frac{97}{120}\)
b.\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
c.\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
\(=\frac{\left(\frac{5}{22}+\frac{3}{13}-\frac{1}{2}\right).\left(2.11.13\right)}{\left(\frac{4}{13}-\frac{2}{11}+\frac{3}{2}\right).\left(2.11.13\right)}\)
\(=\frac{65+66-143}{88-52+429}\)
\(=\frac{-12}{465}=\frac{-4}{155}\)
bài 1 tính
A = \(\frac{2}{3.5}+\frac{2}{5.7}+..............+\frac{2}{37.39}\); B = \(\frac{4}{5.7}+\frac{4}{7.9}+..........+\frac{4}{59.61}\) ; C = \(\frac{4}{5.9}+\frac{4}{9.13}+.................+\frac{4}{41.45}\) Bài 2 chứng minh : \(\frac{m}{b.\left(b+m\right)}=\frac{1}{b}-\frac{1}{b+m}\)
Tính bằng cách hợp lí :
a , \(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{2}\)
b ,\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59\cdot61}\)
c,\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
Tìm a
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right).\)\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+...+\left(a+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Rightarrow12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)(1)
Ta có \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{1}{2}.\frac{24}{25}=\frac{12}{25}\)
Lại có \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)}{2}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{3^5}}{2}=\frac{1}{2}-\frac{1}{3^5.2}\)
Khi đó (1) <=> \(12a-\frac{12}{25}=11a+\frac{1}{2}-\frac{1}{3^5.2}\)
=> \(a=\frac{12}{25}+\frac{1}{2}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{486}=\frac{23764}{24300}\)
Gọi \(A=\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)\)
\(\Rightarrow A=12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{23.25}\right)\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{23.25}\right)\right]\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]\)
\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{25}\right)\right]\)
\(\Rightarrow A=12a+\left(\frac{1}{2}.\frac{24}{25}\right)\)
\(\Rightarrow A=12a+\frac{12}{25}\)
Gọi \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow B=\frac{1}{1.3}+\frac{1}{3.3}+\frac{1}{9.3}+\frac{1}{27.3}+\frac{1}{81.3}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow3B-B=1-\frac{1}{243}\)
\(\Rightarrow2B=\frac{242}{243}\)
\(\Rightarrow B=\frac{121}{243}\)
\(\Rightarrow A=11a+B\)
\(\Rightarrow12a+\frac{12}{25}=11a+\frac{121}{243}\)
\(\Leftrightarrow12a-11a=\frac{121}{243}-\frac{12}{25}\)
\(\Leftrightarrow a=\frac{109}{6075}\)
\(\frac{2}{3}x-\frac{780}{11}:\left(\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\right)=-5\)
2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5
2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5
2\3x-780\11:[13\2.(1\3-1\11)]=-5
2\3x-780\11:[13\2.8\33]=-5
2\3x-780\11:52\33=-5
2\3x-525\13=-5
2\3x=-5+525\13
2\3x=460\13
x=460\13:2\3
x=690\13