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Khách
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23 tháng 11 2016
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
23 tháng 11 2016
(-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
7 tháng 2 2022
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
DT
1
H
7 tháng 5 2019
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
Ta có:\(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75};\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)
Tự giải tiếp hay nhờ thầy cô giảng tiếp đi nha bn, mỏi tay nên ko thể làm đc nữa !!
LM
Lê Minh Vũ
CTVHS
VIP
14 tháng 6 2023
\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)
\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)
= \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)
= \(\dfrac{1}{4}+-\dfrac{6}{5}\)
= \(\dfrac{1}{4}-\dfrac{6}{5}\)
= \(-\dfrac{19}{20}\)
\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)
TA
14 tháng 6 2023
\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)
\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)
\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)
\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)
\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)
\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)
\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)
\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)
\(=-\dfrac{5}{11}\)
\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)
\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)
NT
1
13 tháng 12 2017
a) A= 1-2+3-4+5-6+...+99-100
A = ( 1 - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ... + ( 99 - 100 ) ( có 50 cặp )
A = ( - 1 ) + ( -1 ) + ( -1 ) + ,.. + ( -1 )
A = ( - 1 ) . 50
A = -50
b) B = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + ... + 97 + 98 - 99 - 100
B = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ( 9 + 10 - 11 - 12 ) + ... + ( 97 + 98 - 99 - 100 ) ( có 25 cặp )
B = ( - 4 ) + ( - 4 ) + ( - 4 ) + ... + ( - 4 )
B = ( - 4 ) x 25
B = -100
a) ta có :1/5^2<1/4.5=1/4-1/5
1/6^2<1/5.6=1/5-1/6
.................
1/100^2<1/99.100=1/99-1/100
=>1/5^2+1/6^2+1/7^2+......+1/100^2 <1/4-1/100=6/25<1/4(1)
ta lại có:1/5^2>1/5.6=1/5-1/6
1/6^2>1/6.7=1/6-1/7
.................
1/100^2>1/100.101=1/100-1/101
=>1/5^2+1/6^2+1/7^2+......+1/100^2>1/5-1/101=96/505>1/6(2)
từ (1)(2) suy ra 1/6<1/5^2+1/6^2+1/7^2+......+1/100^2 < 1/4
b)ta có:1/11+1/12+....+1/70=(1/11+1/12+...+1/20)+(1/21+1/22+...+1/30)+(1/31+1/32+...+1/40)+(1/41+1/42+...+1/50)+(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)>(1/20+1/20+...+1/20)(10 phân số 1/20)+(1/30+1/30+...+1/30)(10 phân số 1/30)+(1/40+1/40+...+1/40)(10 phân số 1/40)+(1/50+1/50+...+1/50)(10 phân số 1/50)+(1/60+1/60+...+1/60)(10 phân số 1/60)=1/2+1/3+1/4+1/5+1/6=29/20>4/3(1)
ta lại có:1/11+1/12+....+1/70=(1/11+1/12+...+1/20)+(1/21+1/22+...+1/30)+(1/31+1/32+...+1/40)+(1/41+1/42+...+1/50)+(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)<(1/11+1/11+...+1/11)(10 phân số 1/11)+(1/21+1/21+...+1/21)(10 phân số 1/21)+(1/31+1/31+...+1/31)(10 phân số 1/31)+(1/41+1/41+...+1/41)(10 phân số 1/41)+(1/51+1/51+...+1/51)(10 phân số 1/51)+(1/61+1/61+...+1/61)(10phân số 1/61) =10/11+10/21+10/31+10/41+10/51+10/61=2,311777327<5/2(2)
từ (1)(2)=>4/3<1/11+1/12+....+1/70<5/2