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!!

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

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

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

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

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

\(x\cdot100=5750-5050\)

\(x\cdot100=700\)

\(x=700\div100\)

\(x=7\)

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

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

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

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

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

x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ... + ( x + 100 ) = 11 000

<=> ( x + x + x + x + x + ... + x ) + ( 1 + 2 + 3 + 4 + ... + 100 ) = 11 000

<=> 101x +  \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 11 000 ( vì sao em để 101x thì idol biết mà :33 )

<=> 101x + 5050 = 11 000

<=> 101x = 5950

<=> x = 5950/101

x + (x + 1) + (x + 2) + ......+ (x + 100) = 11000

x +( x + x + ...... + x ) + (1 + 2 + ...... + 100) = 11000

x + 100x + 5050 = 11000

x + 100x = 5950

101x = 5950

x = 5950 : 101

x = 5950 : 100 + 5950 : 1

x = 59,50 + 5950

x = 6009,50

\(x+\left(x+1\right)+\left(x+2\right)+.................+\left(x+100\right)=11000\)

\(x+x+1+x+2+x+3+..............+x+100=11000\)

\(\left(x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=11000\)

Có 101 số hạng x                                         Có 100 số hạng

\(101x+\left[\left(1+100\right).100:2\right]=11000\)

\(101x+\left(101.100:2\right)=11000\)

\(101x+5050=11000\)

\(101x=11000-5050\)

\(101x=5950\)

\(x=\frac{5950}{101}\)

Vậy \(x=\frac{5950}{101}\)

Chúc bạn học tốt

\(\left(x-1\right)\left(-x+2\right)=0\Leftrightarrow x=1;x=2\)

\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)

\(\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\left(x-2\right)\left(-x-2\right)=0\Leftrightarrow x=-2;x=2\)

\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200