A = -1 - 2 - 3 - ... - 100

= -(1 + 2 + 3 + ... + 100)

= -100.101 : 2

= -5050

--------

B = -2 - 4 - 6 - ... - 100

= -(2 + 4 + 6 + ... + 100)

Số số hạng của B:

(100 - 2) : 2 + 1 = 50 (số)

B = -(100 + 2) . 50 : 2 = -2550

--------

C = -6 - 9 - 12 - ... - 99

= -(6 + 9 + 12 + ... + 99)

Số số hạng của C:

(99 - 6) : 3 + 1 = 32 (số)

C = -(99 + 6) . 32 : 2 = -1680

--------

D = 4 - 8 + 12 - 16 + ... + 196 - 200

Số số hạng của D:

(200 - 4) : 4 + 1 = 50 (số)

D = (4 - 8) + (12 - 16) + ... + (196 - 200)

= -4 + (-4) + ... + (-4) (25 số -4)

= -4.25

= -100

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

làm đúng có tick các bạn ơi

\(a,\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{1}{18}=\dfrac{12}{18}+\dfrac{15}{18}+\dfrac{1}{18}=\dfrac{28}{18}=\dfrac{14}{9}\)

\(b,\dfrac{3}{4}+\dfrac{2}{25}+\dfrac{7}{100}=\dfrac{75}{100}+\dfrac{8}{100}+\dfrac{7}{100}=\dfrac{90}{100}=\dfrac{9}{10}\)

\(c,\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{3}{4}=\dfrac{1}{12}+\dfrac{2}{12}+\dfrac{9}{12}=\dfrac{12}{12}=1\)

a)

`2/3 +5/6+1/18`

`=12/18 +15/18 +1/18`

`=28/18`

`=14/9`

b)

`3/4 +2/25+7/100`

`=75/100 +8/100 +7/100`

`=90/100`

`=9/10`

c)

`1/12 +1/6+3/4`

`=1/12+2/12+9/12`

`=12/12`

`=1`

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

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{11}{25}\)

b) \(x\cdot9,85+x\cdot0,15=0,1\)

\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)

\(\Rightarrow x\cdot10=0,1\)

\(\Rightarrow x=\dfrac{0,1}{10}\)

\(\Rightarrow x=0,01\)

c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)

\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)

\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)

\(\Rightarrow2022x=0\)

\(\Rightarrow x=\dfrac{0}{2022}\)

\(\Rightarrow x=0\)

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)

Ta có :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)

\(\Rightarrow x+\dfrac{103}{50}=5\)

\(\Rightarrow x=5-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{147}{50}\)

a) ( 1 - 1/2 ) x ( 1 - 1/3 ) x ( 1 - 1/4 ) x ( 1 - 1/5 ) x ( 1 - 1/6 )

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6 

= \(\frac{1.2.3.4.5}{2.3.4.5.6}\)

= 1/6

\(\frac{1}{6}\)

1/6

đáp số:1/6