Question

Bài 1. Thực hiện phép tính

\(\dfrac{1}{2002}\) + \(\dfrac{2003*2001}{2002}\)+ 2003

Bài 2.Tìm 2 số hữu tỉ a, b biết a+b= a.b = a:b

Bài 3.

Chứng minh rằng: a, \(\dfrac{1}{a(a+1)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)

b, \(\dfrac{2}{a(a+1)(a+2)}=\dfrac{1}{a(a+1)}-\dfrac{1}{(a+1)(a+2)}\)

Answer 1

bài 1\(\dfrac{1}{2002}+\dfrac{2003\cdot2001}{2002}+2003=\dfrac{1+2003\cdot2001+2003\cdot2002}{2002}=\dfrac{1+2003\left(2001+2003\right)}{2002}=1+2003\cdot2=4007\)

Answer 2

câu3

a)VP=\(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}\)=VT

b)VP=VT\(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}=\dfrac{a+2}{a\left(a+1\right)\left(a+2\right)}-\dfrac{a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\dfrac{2}{a\left(a+1\right)\left(a+2\right)}\)

Answer 3

ta có

a+b=ab=>a=ab-b=b(a-1)

Thay a=b(a-1)vào a+b=a/b ta có

\(a+b=\dfrac{b\left(a-1\right)}{b}\Rightarrow b=-1\)thay b=-1 vao a+b=ab ta đc

a-1=-a=>a=1/2