ta có
x+y+y+z+z+x=\(\frac{13}{12}\)
2(x+y+z)=\(\frac{13}{12}\)
=>x+y+z=\(\frac{13}{24}\)
z=(x+y+z)-(x+y)
y=y+z-z
x=x+Y-y
ta có
x+y+y+z+z+x=\(\frac{13}{12}\)
2(x+y+z)=\(\frac{13}{12}\)
=>x+y+z=\(\frac{13}{24}\)
z=(x+y+z)-(x+y)
y=y+z-z
x=x+Y-y
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)\(\frac{1}{95}\)
Tìm x, biết:
\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
Mọi người giúp mình với mình gắp lắm!!!!
like cho những bạn trả lời hợp lý!!!
Bài 1 : Tìm x
1) \(\frac{x-1}{10}\) + \(\frac{x-1}{11}\)+\(\frac{x-1}{12}\)-\(\frac{x-1}{13}\)= 0
2 ) \(\frac{x-1}{99}\)+\(\frac{x-2}{98}\)+\(\frac{x-5}{95}\)= 3 + \(\frac{1}{99}\)+\(\frac{1}{98}\)+\(\frac{1}{95}\)
Tìm x , biết :
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
Tính \(T=\left(\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)X\left(\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}\right)-\left(\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\right)X\left(\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}\right)\)
Tim x,y,z:\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và 2x+3y-z=95
AI GIÚP MÌNH VỚI!!!
1) \(\left(\frac{1}{12}+3\frac{1}{6}-30,75\right)\)\(.x-8\) \(\left(\frac{3}{5}+0,415+\frac{1}{200}\right)\)\(:0,01\)
2) \(\left(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}\right)\)\(.x=\) \(\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}\)
3) \(x-\) \(\frac{20}{11.13}\)\(-\) \(\frac{20}{13.15}\)\(-\)\(\frac{20}{15.17}\)\(-...-\) \(\frac{20}{53.55}=\frac{3}{11}\)
4) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-5}{95}=3+\frac{1}{99}+\frac{1}{98}+\frac{1}{95}\)
GIÚP MÌNH ĐI MÀ!!!😭😭😭
Bài 1 : Tìm x thuộc Z biết : ( Trình bày rõ 2 ý => 2 likes )
a, \(\frac{x+3}{47}+\frac{x+2}{48}=-2\)
b, \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-1\)
bài 1 tìm x :
a) \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
b)\(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)