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\(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.\dfrac{8}{9}.\dfrac{10}{11}.\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
= \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}.\dfrac{7}{8}.\dfrac{8}{9}.\dfrac{9}{10}.\dfrac{10}{11}\)
= \(\dfrac{1}{11}\)
Số nghịch đảo của \(\dfrac{1}{11}\)là \(\dfrac{11}{1}=11\)
a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)
b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)
c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)
d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)
\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)
mà \(10^7-8< 10^8-7\)
nên A>B
c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)
\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)
mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)
nên A<B
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
`#lv`
`A=(-1)+(-5)+(-9)+...+(-101)`
`=-(1+5+9+...+101)`
Số số hạng là :
`[101-(-1)]:4+1=26(` số hạng `)`
Tổng là :
`[(-101)+(-1)]xx26:2=-1326`
Vậy `A=-1326`
__
`B=-5/17 . 8/19 + (-12)/17 . 8/19 - 11/19`
`=((-5)/17+(-12)/17).8/19-11/19`
`=-1.8/19-11/19`
`=-8/19-11/19`
`=-8/19+(-11)/19`
`=-19/19`
`=-1`
__
`C=10/1.6 + 10/6.11 + 10/11.16 + ... + 10/2016.2021`
`=2.(1-1/6+1/6-1/11+...+1/2016-1/2021)`
`=2(1-1/2021)`
`=2. (2021/2021-1/2021)`
`=2. 2020/2021`
`=4040/2021`
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
Cái này mk từng làm nhưng có chút sai sót vậy nên bn sữa cho mk chút nhé ! Thay vì N = ... thì bn thay bằng A = ... nha
Ta có :
N = 40 ( A = 40 )