<=> ( x + x + x + .... + x + x ) + ( 1 + 2 + 3 + .... + 49 + 50 ) = 1425

<=> 50x + { [ 50.( 50 + 1 ) ] : 2 } = 1425

=> 50x + 1275 = 1425

=> 50x = 1425 - 1275

=> 50x = 150

=> x = 150 : 50 = 3

=> (x+ x+ x +...+ x) + (1+2+..+ 50) = 1425

=> 50x + 1275=1425

=> 50x = 150

=> x=3

**** mk nha!

<=>(x+x+...+x)+(1+2+3+...+50)=1425

=>50x+1275=1425

=>50x=1425-1275

=>50x=150

=>x=3

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{2}{9}.\frac{1}{2}\)

\(\Rightarrow\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

\(\frac{1}{x+1}=\frac{1}{18}\Rightarrow18.1=1\left(x+1\right)\)

\(\Rightarrow18=x+1\Rightarrow x=18-1=17\)

Cảm ơn bạn nha!

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\right].x=\frac{23}{45}\)

\(\Leftrightarrow\left(\frac{1}{2}.\frac{44}{90}\right).x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}.x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

nhung sao banj khong phan h ra ro rang,chang nhe den do khong phan h  duoc sao

Ahihi

1.2(x-1)+(x-2)=x-4

         2x-2+x-2=x-4

            2x+ x-x=2+2-4

                    2x=0

=>                  x=0

Vậy x=0

Ta có : \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+1000\right)=5750\)

\(\Leftrightarrow1000x+\left(1+2+...+1000\right)=5750\)

\(\Leftrightarrow1000x+\frac{1001.1000}{2}=5750\)

\(\Leftrightarrow1000x=\frac{5750}{500500}=\frac{23}{2002}\)

\(\Leftrightarrow x=\frac{23}{2002000}\)

(x+1)+(x+2)+...+(x+1000)=5750

=>x.1000+(1+2+3+...+1000)=5750

=>x.1000+[(1000+1).1000:2]=5750

=>1000x+(1001000:2)=5750

=>1000x+500500=5750

=>1000x=5750-500500=-494750

=>x=-494750:1000=-494,75

Có lẽ đúng >: ko chắc :P

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

                   \(x+1+x+2+...+x+100=5750\)phá ngoặc

\(\left(x+x+...+x\right)+\left(1+2+..+100\right)=5750\)ta nhóm lại thành hai nhóm x và 1đến 100

                          \(100x+\left(1+2+..+100\right)=5750\)vì có 100 x nên ta chuyển thành 100x

tách vế     \(1+2+..+100\)

số hạng dãy trên là \(\left(100-1\right):1+1=100\)

tổng dãy trên là\(\left(100+1\right).100:2=5050\)

thay vào\(100x+5050=5750\)ta thực hiện tìm x như bình thường

                               \(100x=700\)

\(\Rightarrow x=700:100=7\)