Đặt A =1/2^2 .1/3^2.1/4^2. ... . 1/99^2

2A=1/2.1/2^2.1/2^3. ... . 1/98^2

2A-A= (1/2.1/2^2.1/2^3. ... . 1^98^2)-(1/2^2.1/3^2.1/4^2. ... . 1/99^2)

A=1/2-1/99^2

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)

\(1-x=\frac{2}{5}:\frac{2}{5}=1\)

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)

\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)

học tốt nha

\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}=\frac{30}{30}=1\)

\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}=\frac{3}{4}-\frac{2}{4}+\frac{2}{5}=\frac{1}{4}+\frac{2}{5}=\frac{5}{20}+\frac{8}{20}=\frac{13}{20}\)

Chúc bạn học tốt!

\(\frac{3}{5}+\frac{1}{6}+\frac{7}{30}\)

\(=\frac{18}{30}+\frac{5}{30}+\frac{7}{30}\)

\(=\frac{18+5+7}{30}\)

\(=\frac{30}{30}\)

\(=1\)

\(\frac{3}{4}-\frac{1}{2}+\frac{2}{5}\)

\(\frac{15}{20}-\frac{10}{20}+\frac{8}{20}\)

\(=\frac{15-10+8}{20}\)

\(=\frac{13}{20}\)

3/5 + 1/6 + 7/30 = 18/30 + 5/30 + 7/30 = 23/30 + 7/30 = 30/30 = 1

3/4 - 1/2 + 2/5 = 3/4 -2/4 + 2/5 = 1/4 + 2/5 = 5/20 + 8/20 = 13/20

Ta có :\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)(1)

\(\Rightarrow3N=1+\frac{1}{3}+...+\frac{1}{3^8}\)(2)

Lấy (2) - (1) ta có :

\(\Rightarrow2N=1-\frac{1}{3^9}\)

\(\Rightarrow N=\frac{1-\frac{1}{3^9}}{2}\)

ta có: \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)

\(\Rightarrow\frac{1}{3}N=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{10}}\)

\(\Rightarrow\frac{1}{3}N-N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(\frac{2}{3}N=\frac{1}{3^{10}}-\frac{1}{3}\)

\(N=\frac{\frac{1}{3^{10}}-\frac{1}{3}}{\frac{2}{3}}\)

CHÚC BN HỌC TỐT!!!

\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^9}\)

\(3N=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^8}\)

\(3N-N=1+\frac{1}{3}+\frac{1}{3^2}+......+\frac{1}{3^8}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^9}\right)\)

\(2N=1+\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^8}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.....-\frac{1}{3^9}\)

\(2N=1-\frac{1}{3^9}\)

\(N=\frac{1-\frac{1}{3^9}}{2}\)

https://olm.vn/hoi-dap/question/893332.html

câu trả lời đó

mk thấy nó cx giống vs bài của cậu

hok tốt

1/1=1 rồi

A=1+2+1+2+3+1+2+3+4+1+2+3+4+5

A=(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)

A=3+6+10+15=24

cách tính 1+2=3+3=6+4=10+5=15(cách đó cho đỡ phải tính kaau)

https://olm.vn/hoi-dap/question/893332.html

câu trả lời đó

hok tốt

Trả lời:

Ta có : 1.2²= 1.2.2=1.2.(3-1)=1.2.3-1.2

             2.3²= 2.3.3=2.3.(4-1)=2.3.4-2.3

.................................................

             98.99²= 98.99.99=98.99.(100-1)=98.99.100-98.99

A=1.2.3 - 1.2 + 2.3.4 - 2.3 + ... + 98.99.100 - 98.99 hay A=1.2.3 + 2.3.4 +...+ 98.99.100 - (1.2 + 2.3 + ... + 98.99) = B - C

B=1.2.3 + 2.3.4 + ... + 98.99.100

B.4=1.2.3.4 + 2.3.4.(5 - 1) + ... + 98.99.100.(101 - 97)= 98.99.100.101

=> 98.99.100.101:4= 24497550

C=1.2 + 2.3 + ... + 98.99 

C.3=1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)= 98.99.100

=> 98.99.100:3= 323400

Vậy A= 24497550 - 323400 = 24174150

@qqqqqqq cảm ơn bạn nhé! (^_^)

-4.|x-1| + (1/2-2,5)² = -3

-4.|x-1| + (-2)² = -3

-4.|x-1| + 4 = -3

-4.|x-1| = -3 - 4

-4.|x-1| = -7

|x-1| = (-7) : (-4)

|x-1| = 7/4

TH1: x - 1 = 7/4 => x = 7/4 + 1 = 11/4

TH2: 1 - x = 7/4 => x = 1 - 7/4 = -3/4

Vậy x = {11/4; -3/4}

= 1

tick đi mink giải thích cho .  hihihihihihihihiihihiiiiiiiiiiiiiiii

A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)  + \(\dfrac{1}{4}-\dfrac{1}{5}\)+ \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)

A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)

A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)

A = 1