Tính:
(1-1/9). (1-2/9). (1-3/9)... (1-2005/9)
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FB
1
S
22 tháng 2 2019
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)\left(1-\frac{3}{9}\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{9}{9}\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...\left(1-1\right)...\left(1-\frac{2005}{9}\right)\)
\(I=\left(1-\frac{1}{9}\right)\left(1-\frac{2}{9}\right)...0...\left(1-\frac{2005}{9}\right)\)
I = 0
=> I = 0
KH
0
11 tháng 8 2017
a, \(\frac{4}{7}x\frac{5}{7}+\frac{3}{7}x\frac{5}{6}\)
= ( \(\frac{4}{7}+\frac{3}{7}\)) x \(\frac{5}{7}x\frac{5}{6}\)
= 1 x \(\frac{5}{7}x\frac{5}{6}\)
= \(\frac{5}{7}x\frac{5}{6}\)
= \(\frac{25}{42}\)
Tương tự mấy câu sau cũng làm như thế này
9 tháng 8 2019
\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)
\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=>1+\frac{16}{21}-\frac{20}{15}\)
\(=>\frac{37}{21}-\frac{20}{15}\)
\(=>\frac{3}{7}\)
\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)
\(=>12-8\cdot\frac{27}{8}\)
\(=>12-27\)
\(=>-15\)
\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)
\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)
\(=>\left(1^{2005}\right)-16\)
\(=>1-16\)
\(=>-15\)
22 tháng 10 2019
a)=(3/8+10/16)+(7/12+10/24)
=1+1=2
c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)
=3+2+2=7
PT
1
2 tháng 3 2019
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times...1\frac{1}{9}+2005.\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times...\times\frac{10}{9}+2005\)
\(=\frac{10}{2}+2005=5+2005=2010\)
.
25 tháng 9 2019
a) 1+2+3+4+5+...+n = n(n+1) / 2
b)2+4+6+...+2n = [(2n-2):2+1] . (2n+2)/2 = n . ( 2n+2) /2
\(\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{9}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{2}{9}\right)\cdot...\cdot0\cdot...\cdot\left(1-\dfrac{2005}{9}\right)=0\)
\(\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{9}{9}\right)...\left(1-\dfrac{2005}{9}\right)\)
\(=\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{2}{9}\right)...0...\left(1-\dfrac{2005}{9}\right)=0\)