\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\left(x\cdot100\right)+\left(1+2+...+100\right)=5750\)

\(\left(x\cdot100\right)+\left(100+1\right)\cdot\frac{100}{2}=5750\)

\(\left(x\cdot100\right)+101\cdot50=5750\)

\(\left(x\cdot100\right)+5050=5750\)

\(x\cdot100=5750-5050\)

\(x\cdot100=700\)

\(x=700\div100\)

\(x=7\)

Ta có: ( x+1)+(x+2)+(x+3)+.....+(x+99)+(x+100)=5750

<=>(x+x+x+....+x+x)+(1+2+3+..+99+100)=5750

<=> 100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

Vậy x=7

 hoặc mở câu hỏi tương tự tham khảo.

thank you

=(x+x+...+x)+(1+2+...+100)=5750

=100x+5050=5750

           100x=5750-5050

           100x=700

               x=700/100=7

(x+1)+(x+2)+....+(x+100)=5750

x+1+x+2+...+x+100       =5750

\(x\)x100+1+2+...+100     =5750

         bí  hihi 

tự làm nha 

( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 5750

x \(\times\) 100 + 1 + 2 + ... + 100 = 5750

x \(\times\) 100 + 5050 = 5750

x \(\times\) 100 = 700

x  = 7

\(\left(x+1\right)+\left(x+2\right)+.....+\left(x+100\right)=5750\)

\(\Rightarrow x+1+x+2+.....+x+100=5750\)

\(\Rightarrow100x+1+2+3+....+100=5750\)

\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)

\(\Rightarrow100x+5050=5750\)

\(\Rightarrow100x=700\)

\(\Rightarrow x=7\)

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+...+100\right)=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(x=7\)

Vậy ...

100x+5050=5750

100x=5750-5050

100x=700

x=700:100

x=7

\(a)\) \(A=4+2^2+2^3+...+2^{20}\)

\(A=2^2+2^2+2^3+...+2^{20}\)

\(2A=2^3+2^3+2^4+...+2^{21}\)

\(2A-A=\left(2^3+2^3+2^4+...+2^{21}\right)-\left(2^2+2^2+2^3+...+2^{20}\right)\)

\(A=2^3+2^{21}-2^2-2^2\)

\(A=2^3+2^{21}-2.2^2\)

\(A=2^3+2^{21}-2^3\)

\(A=2^{21}\)

Vậy \(A=2^{21}\)

\(b)\) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow\)\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(\Leftrightarrow\)\(100x+\frac{100\left(100+1\right)}{2}=5750\)

\(\Leftrightarrow\)\(100x+5050=5750\)

\(\Leftrightarrow\)\(100x=5750-5050\)

\(\Leftrightarrow\)\(100x=700\)

\(\Leftrightarrow\)\(x=\frac{700}{100}\)

\(\Leftrightarrow\)\(x=7\)

Vậy \(x=7\)

Chúc bạn học tốt ~ 

b,  (x+x+x+....+x)+(1+2+3+4+...+100)=5750

    100x+5050=5750

    100x=5750-5050

    100x=700

     x=700/100

     x=7

a, Vì trong mỗi ngoặc có một số hạng nên vì có 100 số hạng nên có 100x

ta có 100x+(1+2+3+.....+100)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=700:100

x=7 

nếu tính ko nhầm

a)(x+1)+(x+2)+...+(x+100)=5750

(x+x+...+x)+(1+2+...+100)=5750

1+2+...+100 có: (100-1)+1 =100 số hạng

1+2+...+100=(100+1)*100/2=5050

=>100x+5050=5750

100x=5750-5050

100x=700

x=700/100

x=7. Vậy x=7

Ta có:100x+(1+2+3+...+100)=5750

=>100x+5050=5750

=>100x=5750-5050

=>100x=700

=>x=700:100

=>x=7

Vậy x=7

c) (x+1) + (x+2) + ... + (x+5) = 90

=> 5x + ( 1 + 2 + ... + 5 ) = 90

5x + 15 = 90

5x = 90 - 15

5x = 75

x = 75 : 5

x  = 15

d) (x+1) + (x+2) + .... + (x+100) = 20150

=> 100x + ( 1+2+...+100 ) = 20150

100x + 5050 = 20150

100x = 20150 - 5050

100x = 15100

x = 15100 : 100

x = 151

Ta có : (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 90

<=> x + x + x+ x + x + (1 + 2 + 3 + 4 + 5) = 90

<=> 5x + 15 = 90

=> 5x = 75

=> x = 15 

c)   \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=90\)

 \(\Leftrightarrow x+1+x+2+x+3+x+4+x+5=90\)

\(\Leftrightarrow5x+\left(1+2+3+4+5\right)=90\)

\(\Leftrightarrow5x+15=90\)

\(\Leftrightarrow5x=75\)

\(\Leftrightarrow x=15\)

d)    \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+......+\left(x+99\right)+\left(x+100\right)=20150\)

\(\Leftrightarrow x+1+x+2+x+3+......+x+99+x+100=20150\)

\(\Leftrightarrow100x+\left(1+2+3+.....+99+100\right)=20150\)

\(\Leftrightarrow100x+5050=20150\)

\(\Leftrightarrow100x=15100\)

\(\Leftrightarrow x=151\)

thực hiện phép tính

Phân thức cuối hình như mẫu sai rồi bạn

Phải là (x+9)(x+10) mới đúng chứ

Nếu đề bài đúng thì sẽ làm như sau:

A = \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\) (ĐKXĐ tự tìm nhé, chứ viết dài lắm)

A = \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+9}-\frac{1}{x+100}\)

A = \(\frac{1}{x}-\frac{1}{x+100}\)

A = \(\frac{x+100-x}{x\left(x+100\right)}\)

A = \(\frac{100}{x\left(x+100\right)}\)

Vậy A = \(\frac{100}{x\left(x+100\right)}\)

Chúc bn học tốt!!

        \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+.....+\frac{1}{\left(x+99\right)\left(x+100\right)}\)

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

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

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{1}{x}-\frac{1}{x+100}=\frac{x+100-x}{x\left(x+100\right)}=\frac{100}{x\left(x+100\right)}\)