a)x² - y² - 2x + 2y=(x^2-y^2)-(2x+2y)=(x-y)(x+y)-2(x+y)=(x-y)(x+y-2)

b)2x + 2y - x² - xy=(2x+2y)-(x^2-xy)=2(x+y)-x(x+y)=(2-x)(x+y)

c,d bạn làm tương tự nha^^

Vì \(F\left(x\right)\)chia \(x^2+4x+3\)được thương là \(2017\)và có dư nên \(F\left(x\right)\)có dạng: 

\(F\left(x\right)=2017\left(x^2+4x+3\right)+ax+b\)

\(F\left(x\right)\)chia \(x+1\)dư \(3\)nên \(F\left(-1\right)=3\)

\(F\left(x\right)\)chia \(x+3\)dư \(-1\)nên \(F\left(-3\right)=-1\)

Có: \(\hept{\begin{cases}F\left(-1\right)=-a+b=3\\F\left(-3\right)=-3a+b=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=5\end{cases}}\)

Vậy \(F\left(x\right)=2017\left(x^2+4x+3\right)+2x+5\).

\(=\frac{\left[\left(1502-1\right)\left(1502+1\right)\right]\left[\left(1499+1\right)\left(1499-1\right)\right]}{6002}\)

\(=\frac{\left(1502^2-1\right)-\left(1499^2-1\right)}{6002}=\frac{2256003-2247000}{6002}=\frac{9003}{6002}=1,5\)

\(D=\frac{1501.1503-1500.1498}{6002}\)

\(\Leftrightarrow D=\frac{\left(1502-1\right)\left(1502+1\right)-\left(1499+1\right)\left(1499-1\right)}{6002}\)

\(\Leftrightarrow D=\frac{1502^2-1-1499^2+1}{6002}\)

\(\Leftrightarrow D=\frac{\left(1502-1499\right)\left(1502+1499\right)}{6002}\)

\(\Leftrightarrow D=\frac{3.3001}{6002}=\frac{9003}{6002}=\frac{3}{2}\)

\(C=\frac{1999.4001+2000}{2000.4001-2001}\)

\(\Leftrightarrow C=\frac{\left(2000-1\right).4001+2000}{2000.4001-2001}\)

\(\Leftrightarrow C=\frac{2000.4001-4001+2000}{2000.4001-2001}\)

\(\Leftrightarrow C=\frac{2000.4001-2001}{2000.4001-2001}=1\)

 \(D=\frac{1501.1503-1500.1498}{6002}\)

\(\Leftrightarrow D=\frac{\left(1502-1\right)\left(1502+1\right)-\left(1499+1\right)\left(1498-1\right)}{6002}\)

\(\Leftrightarrow D=\frac{1502^2-1-1499^2+1}{6002}\)

\(\Leftrightarrow D=\frac{\left(1502-1499\right)\left(1502+1499\right)}{6002}\)

\(\Leftrightarrow D=\frac{3.3001}{6002}=\frac{3.3001}{2.3001}=\frac{3}{2}\)

 So sánh 1 và 3/2, Ta thấy C<D

\(C=\frac{\left(3000-1001\right)\left(3000+1001\right)+2000}{2000\left(4001-1\right)+1}\)

\(=\frac{3000^2-1001^2+2000}{2000\cdot4000+1}=\frac{9000000-1002001+2000}{8000000+1}\)

\(=\frac{7999999}{8000001}=0,99999975\)

còn câu D mk làm ở câu khác rồi nên mk ghi luôn kết quả nha

D =  1,5

=>  C <  D 

\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow-2\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)=0\Leftrightarrow x=\frac{5}{2}\)