Tính:
A= (1/10-1) . (1/11-1) . (1/12-1) . ... . (1/99-1) . (1/100-1)
Giúp mình nha ! Thanks !
CHO \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\)
SO SÁNH A VỚI 1 GIÚP MÌNH GIẢI ĐẦY ĐỦ NHÉ ! THANKS
A= 1/10+1/11+1/12+1/13+...........+1/99+1/100
2A=1/9+1/10+1/11+1/12+...........+1/98+1/99
2A-A=(1/10+1/11+1/12+1/13+.............+1/99+1/100)-(1/9+1/10+1/11+1/12+............1/98+1/99)
A=1/100-1/9
Tính:A=1+2-3-4+5+6-7-8+9+10-11-12+.....+97+98-99-100
\(\text{A = 1-2-3-4+5-6-7-8+9-10-11-12+...........+97-98-99-100}\)
\(\text{A =(1-2-3-4)+(5-6-7-8)+(9-10-11-12)+.............+(97-98-99-100)}\)
\(\text{A =-8+(-16)+(-24)+..................+(-200)}\)
\(\text{A =-8.(1+2+3+......+25)}\)
\(\text{A =-8.[(25-1):1+1.26:2]}\)
\(\text{A =-8.325}\)
\(\text{A =-2600 Vậy A = -2600 }\)
=1+(2-3-4+5)+(6-7-8+9)+...+(98-99-100+101)+102
=1+0+0+0+....+102
=103
k mk nha
ne bang 153 dung khong nhi
153 ddddo
A= 1/1×2+1/2×3+...1/98×99+1/99×100
B=4/3×7+4/7×11+4/11×15+...4/107×111
C=7/10×11+7/11×12+7/12×13+...7/69×70
Các bạn làm ơn giúp mình với
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}\)
\(\Rightarrow C=\frac{3}{5}\)
A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{99}+\frac{1}{100}\).CHỨNG TỎ A >1
MÌNH GẤP LẮM GIÚP MÌNH VỚI
ta có : \(\frac{1}{10}>\frac{1}{100}\)
\(\frac{1}{11}>\frac{1}{100}\)
\(\frac{1}{12}>\frac{1}{100}\)
\(..............\)
\(\frac{1}{99}>\frac{1}{100}\)
\(\frac{1}{100}=\frac{1}{100}\)
cộng vế với vế ta được :
\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>\frac{91}{100}>1\)
hiệp sai rùi
vầy nè : ta tách a thành 2 nhóm
nhom1 tu 1/10 den 1/50 ta dat =b
ta có :b=1/10+1/11+1/12+.......1/50
b>41/50 (vi 1/10 >1/50;1/11>1/50;....1/50=1/50)
nhóm 2 =c làm tương tự >50/100
a= b+c>50/100+41/50=33/25>25/25=1
mình ko giải chi tiết đâu bạn
Ta có: A = \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{10}>\frac{1}{100};\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};....;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow A>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(\Rightarrow A>\frac{1}{10}+\frac{90}{100}=\frac{100}{100}=1\) (ĐPCM)
1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)+11+(-12)=
-1+2+(-3)+4+(-5)+6+(-7)+8+(-9)+10+(-11)+12=
(-1)+(-2)+(-3)+(-4)+.......+(-99)+(-100)=
(-1)+2+(-3)+4+.......+(-99)+100=
1+(-2)+3+(-4)+........+99+(-100)=
lam la co tick nha
1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)+11+(-12)
=(1+3+5+7+9+11)+[(-2)+(-4)+(-6)+(-8)+(-10)+(-12)]
= 36+-42
=-6
(-1)+2+(-3)+4+(-5)+6+(-7)+8+(-9)+10+(-11)+12
=[(-1)+(-3)+(-5)+(-7)+(-9)+(-11)]+(2+4+6+8+10+12)
=(-36)+42
=6
cho A=1/10+1/11+1/12+...+1/99+1/100. chung minh A> 1 , giup mk voi nha
Tính A = (1/10-1)(1/11-1)(1/12-1)....(1/99-1)(1/100-1)
Giải giúp mình vs ạ 🙂❤️❤️
CẦN GẤP 😭😭😭😭😭😭
Xin á 😂😂😂😂
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(=\frac{-9}{10}.\frac{-10}{11}.\frac{-11}{12}...\frac{-98}{99}.\frac{-99}{100}\)
\(=-\frac{9.10.11....98.99}{10.11.12...99.100}=-\frac{9}{100}\)
chứng tỏ rằng B= 1/10+1/11+1/12+....+1/99+1/100 B<1 nhanh giúp mk mai thi rồi
x=1+2+3-4-5-6+7+8+9-10-11-12+...+97+98+99-100-101-102
Giúp mình nha.
Ai nhanh mình tick
x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)
x=-9+...+-9
x=-9.17
x=-153