a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0

=> x + 100 = 0 => x = -100

b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)

\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)

\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)

Vì 1/47 + 1/48 - 1/49 khác 0

Nên x -50 = 0 => x = 50

giúp mình với ạ

--- là dấu trừ ạ

\(\frac{x}{0,3}=\frac{y}{0,7}=z=\frac{3x}{0,9}\)\(=\frac{z-3x}{1-0,9}=\frac{1}{0,1}=10\)( tính chất dãy tỉ số bằng nhau )

\(=>\hept{\begin{cases}x=0,3.10=3\\y=0,7.10=7\\z=10\end{cases}}\)

Vậy x = 3; y = 7; z = 1

A= (\(\left(\frac{19^{2010}}{19}+\frac{5^{2010}}{5}\right)^{2010}\)=\(\frac{\left(5.19^{2010}+19.5^{2010}\right)^{2010}}{19^{2010}.5^{2010}}\)= A(1)/A(2)

B = \(\frac{\left(19^{2010}+5^{2010}\right)^{2010}}{19^{2010}+5^{2010}}\)= B(1)/B(2)

Ta thấy A(1) >B(1), A(2)<B(2) => A>B

ủa bạn duchinhle tại sao 19^2010.5^2010 lại lớn hơn 19^2020+5^2010

Vì góc COD,AOB là góc bẹt nên =180 độ 

Góc AOD = (180-20):2=80

AOC =180-80=100

COB=180-100=80

BOD=180-80=100

mk cx k chắc lắm nha

Ta có1/51<1/50  ,  1/52<1/50......1/60<1/50

=>1/51+1/51+...+1/60< 1/50.10

=>1/51+1/51+...+1/60<1/5   ,1/5<1/2

=> 1/51+1/51+...+1/60<1/2

=>S<1/2

Rút gọn niểu thức sau :

A= -52 - 5 . 3 / 53 + 52 . 32

B= 46  . 95 + 69 . 120 / 84 . 312 - 611