\(\text{Ta có: a,(x+1)+(x+3)+(x+5)+...+(x+99)=0 }\)

\(\Leftrightarrow50x+\left(1+3+5+...+99\right)=0\)

\(\Leftrightarrow50x+2500=0\)

\(\Rightarrow50x=-2500\)

\(\Rightarrow x=-2500:50=-50\)

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

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

Dấu chấm là nhân

a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)

CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow x+100=0\)

hay x=-100

Mn nhớ giải chi tiết ra nha

ai nhanh nhất mình k cho

Bài 1:

\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)

\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)

\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)

\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)

\(A=\frac{9}{11}\)

Bài 2 :

2)  Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15

= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15

Mẫu số =  11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19  + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17  

=  (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x  17 = (1+27+125+729) x 13 x 15 x 17 

\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)

\(=\frac{11}{17}\)

\(=\frac{1111}{1717}=B\)

Vậy \(A=B\)

mình đâu biết

a/ 99x+(1+2+...+99)=0

99x+(1+99)x49,5=0

x=5

b/3x-1-2-3+1+2+3+4+...+10=0

3x+4+5+...+10=0

x=-49/3

( 15 - x ) + ( x - 12 ) = ( -5 + x )

( 15 - 12 ) - x + x = -5 + x

3 = -5 + x

x = 3 + 5

x = 8

(x+1)+(x+3)+(x+5)+...+(x+99)=0

( x + x + x + .... + x ) + ( 1 + 3 + 5 + ... + 99 ) = 0

50x + 2500 = 0

50x = -2500

x = -2500 : 50

x = -50

trả lời hết lun đi

a,(x+x+x+...+x)+(1+3++5+...+99)=0

   có 50 số            có 50 số

50x+(99+1)*50/2=0

50x+2500=0

50x=0-2500

50x=-2500

x=-2500/50

x=-50

Vậy x=-50

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1