\(\text{a)}\frac{-1}{5}< 0< \frac{1}{1000}\Rightarrow\frac{-1}{5}< \frac{1}{1000}\)

\(\text{b)}\frac{-267}{268}>-1>\frac{-1347}{1343}\Rightarrow\frac{267}{268}>\frac{-1347}{1343}\)

\(\text{c)}\frac{-13}{38}>\frac{-13}{39}=\frac{-1}{3}=\frac{29}{-87}>\frac{29}{-88}\Rightarrow\frac{-13}{38}>\frac{29}{-88}\)

\(\text{d)}\frac{-18}{31}=\frac{\left(-18\right).10101}{31.10101}=\frac{-181818}{313131}\Rightarrow\frac{-18}{31}=\frac{-181818}{313131}\)

Sai chỗ nào T.T

\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\) 

\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\) 

\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\) 

\(=-1\left(1-\frac{1}{2003}\right)\) 

\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)              

\(=-1\cdot\frac{2002}{2003}\) 

\(=-\frac{2002}{2003}\)

\(=\frac{-1\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}+2}-\frac{-1\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\) 

\(=-1-\left(-1\right)\) 

\(=-1+1\) 

\(=0\)

\(=\frac{-\left(\frac{2}{3}+\frac{3}{4}-2\right)}{\frac{2}{3}+\frac{3}{4}-2}-\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)

\(=\left(-1\right)-\left(-1\right)\)

\(=0\)

O₂ = 55⁰

O₃ = 125⁰

O₄ = 55⁰

cảm ơn bạn

B4:

a)-2/15 - x = -3/10                           

               x =-2/5 - (-3/10)

               x =1/6

b)x -1/15 = 1/10

   x           = 1/10 + 1/15

   x           = 1/6 (0,167)

c)-3/8 - x = 5/12 + 2/3

   -3/8 - x = 12/13

            x = -135/104 (-1,298....)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)

\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)\left(-\frac{3}{4}\right)...\left(-\frac{1998}{1999}\right)\)

\(=\frac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-1998\right)}{2\cdot3\cdot4\cdot...\cdot1999}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot1998}{2\cdot3\cdot4\cdot...\cdot1999}=\frac{1}{1999}\)

Dễ thôi

\(x^2+xy-3x-3y+7=0\)

\(\Leftrightarrow x\left(x+y\right)-3\left(x+y\right)=-7\)

\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=-7=1.\left(-7\right)=\left(-1\right).7\)

Xoq xét các TH là ra

\(x^2+xy-3x-3y+7=0\)

\(\Leftrightarrow\left(x^2+xy\right)-\left(3x+3y\right)=-7\)

\(\Leftrightarrow x\left(x+y\right)-3\left(x+y\right)=-7\)

\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=-7\)

Lập bảng giá trị ta có: 

Vậy các cặp giá trị \(\left(x;y\right)\)thỏa mãn đề bài là \(\left(-4;5\right)\), \(\left(2;5\right)\); \(\left(4;-11\right)\), \(\left(10;-11\right)\)

bạn có thể nói rõ hơn đc ko