a: Số số hạng là:

50-1+1=50(số)

Tổng của dãy là:

\(\dfrac{\left(50+1\right)\cdot50}{2}=1275\)

b: Số số hạng là:

(100-2):2+1=50(số)

Tổng của dãy là:

\(\dfrac{\left(100+2\right)\cdot50}{2}=2550\)

a)A=(50+1).50:2=51.50:2=1275

b)B=(100+2).((100-2):2+1):2=98.50:2=2450

ta có : (ghi lại đề)

=6+12+18+24+30/3+6+9+12+15

=2*(3/3+6/6+9/9+12/12+15/15)

=2*(1+1+1+1+1)

=2*5=10

chúc main học tốt nhé

hi

a)

`1/3+3/4+2/3+1/4`

`=1/3+2/3+3/4+1/4`

`=1+1`

`=2`

b)

`3/4+3/5+2/8+4/10`

`=3/4+2/8+3/5+4/10`

`=6/8+2/8+6/10+4/10`

`=1+1`

`=2`

c)

`1/10+2/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10`

`=1/10+9/10+2/10+8/10+3/10+7/10+6/10+4/10+5/10`

`=1+1+1+1+5/10`

`=4+5/10`

`=40/10+5/10`

`=45/10=9/2`

a: =1/3+2/3+3/4+1/4

=1+1

=2

b: =3/4+1/4+3/5+2/5

=1+1

=2

c: =(1+2+3+4+5+6+7+8+9)/10

=45/10

=9/2

Ai giải giùm mình với

Mình đang cần gấp

a: =4(3h15'+2h4')=4(195'+124')=1276'

b: =3(15'45s-6'45s)=3x9'=27'

Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{8}{9}\)

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

1.

Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}$

$=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1$

$\Rightarrow a=2(-1)=-2; b=3(-1)=-3; c=4(-1)=-4$

2.

$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}$
$=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)

b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).

Vậy : \(S=\dfrac{99}{100}.\)

a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)

\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)

b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

trl sẽ có like

a=8x3x6x4x(427+573)

a=8x3x6x4x1000

a=576000

\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)

\(=1-\left(\dfrac{3}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)

\(=1-\left(\dfrac{7}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)

\(=1-\left(\dfrac{15}{16}+\dfrac{1}{32}\right)\)

\(=1-\dfrac{31}{32}=\dfrac{1}{32}\)