1a. \(\frac{-7}{3}.\frac{4}{9}+\frac{-7}{3}.\frac{5}{9}b\frac{1}{3}+\frac{1}{3^2}+....\frac{1}{3^{99}}\)
B=\(\frac{1}{3}-\frac{3}{4}-\left(-0,6\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
C=\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+0,6-\frac{1}{3}\)
D=\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}.....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)
=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)
=\(\frac{1}{99}-1-\frac{1}{99}\)
=1
B = \(\frac{(\frac{1}{5}-\frac{2}{7})\times\frac{3}{4}-\frac{3}{4}\times(\frac{1}{3}-\frac{2}{7})}{\frac{1}{5}\times\frac{2}{7}-\frac{1}{3}\times(\frac{2}{7}+\frac{3}{9})+\frac{3}{9}\times\frac{1}{5}}\)
ồ cuk dễ nhỉ
Nếu các bn thích thì ...........
cứ cho NTN này nhé !
thực hiện phép tính :
a, \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-58+5+0,7}\)
b, \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
c, \(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
Mong các bạn giúp đỡ nhé
\(\left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right)\)\(\frac{9}{7}\))
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(\frac{1}{2014}-\frac{1}{2014.2013}-\frac{1}{2013.2012}-\frac{1}{2012.2011}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)\(\frac{1}{15}\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
Tính hợp lí:
A =\((\frac{2}{7}\times\frac{1}{4}-\frac{1}{3}\times\frac{2}{7})\div(\frac{2}{7}\times\frac{3}{9}-\frac{2}{7}\times\frac{2}{5})\)
B = \(\frac{(\frac{1}{5}-\frac{2}{7})\times\frac{3}{4}-\frac{3}{4}\times(\frac{1}{3}-\frac{2}{7})}{\frac{1}{5}\times\frac{2}{7}-\frac{1}{3}\times(\frac{2}{7}+\frac{3}{9})+\frac{3}{9}\times\frac{1}{5}}\)
CÓ LỜI GIẢI THÍCH CHI TIETS NHÉ AI NHANH MK TICK
A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
=(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)
Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
TÍNH NHANH:
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
A= 1/3- 3/4+ 3/5+ 1/72- 2/9- 1/36+ 1/15
A= ( 1/3- 3/5+ 1/15) - (3/4- 1/72+ 2/9+ 1/36)
A= (5/15- 9/15+ 1/15) - (54/72- 1/72+ 16/72+ 2/36)
A= 1- 71/72
A= 1/72
A = \(\frac{1}{3}\)- \(\frac{3}{4}\) - ( - \(\frac{3}{5}\)) + \(\frac{1}{72}\)- \(\frac{2}{9}\)- \(\frac{1}{36}\)+ \(\frac{1}{15}\)
A = ( \(\frac{1}{3}\)+ \(\frac{3}{5}\)+ \(\frac{1}{15}\)) - ( \(\frac{3}{4}\)- \(\frac{1}{72}\)+ \(\frac{2}{9}\)+ \(\frac{1}{36}\))
A = ( \(\frac{5}{15}\)+ \(\frac{9}{15}\)+ \(\frac{1}{15}\)) - ( \(\frac{54}{72}\)- \(\frac{1}{72}\)+ \(\frac{16}{72}\)+ \(\frac{2}{72}\))
A = 1 - \(\frac{71}{72}\)
A = \(\frac{1}{72}\)