a) ( 1 + 3 + 5 + ....+ 97 + 99 ) - 2 x X = 500 

Áp dụng công thức tính dãy số ta có :

\(1+3+5+...+99=\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=50.100:2=50.50=2500\)

=> 2500 - 2X = 500

=> 2X = 2500 - 500 = 2000

=> X = 2000 : 2 = 1000

a) Ta có : 1 + 3 + 5 + ..... + 97 + 99 

Số số hạng trên là :

 ( 99 - 1 ) : 2 + 1 = 50 số hạng :

 Tổng trên là :

 ( 99 - 1 ) x 50 : 2 = 2450 

Thế vào câu a ta được :

 2450 - 2x = 500

=> 2x = 1950

=> x = 975

b)  X - 1234 = ( 175 x 2 - 50 x 7 ) : 23 + 23 

=> X - 1234 = (350 - 350) : 23 + 23

=> X - 1234 = 0 : 23 + 23

=> X - 1234 = 0 + 23 = 23

=> X = 23 + 1234

=> X = 1257

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)

\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)

\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)

\(\Leftrightarrow x=23\)

vậy................

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)

\(\Leftrightarrow x=300\)

vậy..........

\(x+\left(x+1\right)+\left(x+12\right)+\left(x+23\right)+\left(x+77\right)+\left(x+88\right)+\left(x+99\right)=370\)

\(\Leftrightarrow x+x+1+x+12+x+23+x+77+x+88+x+99=370\)

\(\Leftrightarrow7x+\left(1+12+23+77+88+99\right)=370\)

\(\Leftrightarrow7x+\left(1+99\right)+\left(12+88\right)+\left(23+77\right)=370\)

\(\Leftrightarrow7x+100+100+100=370\)

\(\Leftrightarrow7x+300=370\)

\(\Leftrightarrow7x=70\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\)

\(x+x+x+x+x+x+x+1+99+12+88+23+77=370\)                                     

\(7x+100+100+100=370\)         

\(7x+300=370\)   

\(7x=70\)    

\(x=10\)

câu b nek:

b) (-19).17+(-19).29+(-19).54

=(-19).(17+29+54)

=(-19).100

=-1900

câu c mk chịu hoi, mk chỉ có kết quả là : -39690

câu d đây:

d, (-23).37+(-23).19+(-23).43+(-23)

=(-23).(37+19+43+1)

=(-23).100

=-2300

Ai đúng mình like !

x=1

x=100

\(x+23=24\)

\(x=24-23\)

\(x=1\)

\(x+99=199\)

\(x=199-99\)

\(x=100\)

x+23=24

x      =24-23

x      =1

x+99=199

x      =199-99

x      =100

a)

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)

$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$

Do đó $x-23=0\Rightarrow x=23$

b)

PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$

$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$

$\Rightarrow x+100=0\Rightarrow x=-100$

c)

PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$

Do đó $x+2005=0\Rightarrow x=-2005$

d)

PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)

\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)

\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$

Do đó $300-x=0\Rightarrow x=300$