Mình sửa lại để nha \(\left(125^7-625^5-25^9\right)\div99\)

Ta có \(\frac{\left(5^3\right)^7-\left(5^4\right)^6-\left(5^2\right)^9}{99}\)

\(=\frac{5^{21}-5^{20}-5^{18}}{99}\)

\(=\frac{5^{18}.\left(5^3-5^2-1\right)}{99}\)

\(=\frac{5^{18}.99}{99}\)

\(=5^{18}\)

a, 

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)

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

\(=\frac{1}{2}\times\frac{x+1}{x+2}\)

\(=\frac{2x+2}{x+2}\)

Hơ hơ =v

Làm đại phần a đúng sai mặc kệ ~~

a,

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x\left(x+2\right)}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\left(x+2\right)}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)

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

\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)

\(=\frac{2x+2}{x+2}\)

b,

x = 1.2 + 2.3 + 3.4 + ....+ 89.90

3x = 1.2.3 + 2.3.3 + 3.4.3 + .... + 89.90

3x = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 89.90.(91 - 88)

3x = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 89.90.91 - 88.89.90

3x = 89.90.91

x = \(\frac{89\cdot90\cdot91}{3}=242970\)

Ta có \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=50x\)

\(\Rightarrow100x+\left(1+2+3+....+99\right)=50x\)

\(\Rightarrow100x+4950=50x\)

\(\Rightarrow-50x=4950\)

\(\Rightarrow x=-99\)

x + x + 1 + ... + x + 99 = 50x

Có số số hạng là : ( 99 - 1 ) : 1 + 1 = 99 ( số )

Tổng là : ( 99 + 1 ) . 99 : 2 = 4950

=> 100x + 4950 = 50x

        4950          = -50x

=> x = -99

Vậy,........

T có : 2n - 4 = n + 2 + n + 2 - 8

Vì n + 2 chia hết cho n + 2 

=> -8 chia hết cho n + 2 

=> n + 2 thuộc Ư ( 8 ) = { 1 ; 2 ; 4 ; 8 ; -1 ; -2 ; -4 ; -8 }

Chia Trường hợp ra làm nốt nhé =v

= 5/2

K nha

bạn giải hộ mình với

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2017}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\)

\(A=\frac{1}{2017}\)

\(\frac{1-1}{2}.\frac{1-1}{3}.\frac{1-1}{4}......\frac{1-1}{2017}.\frac{1-1}{2018}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}........\frac{2016}{2017}.\frac{2017}{2018}\)

\(=\frac{1}{2018}\)

Ta có \(\frac{27}{3}-\left(x-\frac{1}{2}\right)=\frac{120}{160}\)

\(\Rightarrow9-\left(x-\frac{1}{2}\right)=\frac{3}{4}\)

\(\Rightarrow x-\frac{1}{2}=9-\frac{3}{4}\)

\(\Rightarrow x-\frac{1}{2}=\frac{33}{4}\)

\(\Rightarrow x=\frac{35}{4}\)

27/3 -(x-1/2)=120/160

9-(x-1/2)=3/4

x-1/2=9-3/4

x-1/2=33/4

x=33/4+1/2

x=35/4

vậy x= 35/4

 k cho mk nha

s=11025    s trồng rau =2205