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Ta có : P = \(\dfrac{1}{1975}\left(\dfrac{2}{1945}-1\right)-\dfrac{1}{1945}\left(1-\dfrac{2}{1975}\right)+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2}{1975.1945}-\dfrac{1}{1975}-\dfrac{1}{1945}+\dfrac{2}{1975.1945}+\dfrac{1974}{1975}.\dfrac{1946}{1945}\)
\(-\dfrac{3}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{2+2+1974.1946-3-1975-1945}{1975.1945}\)
= \(\dfrac{1973}{1975}\)
a) \(A=85^2-45^2+75^2-35^2+65^2-25^2+55^2-15^2\)
\(A=\left(85-45\right)\left(85+45\right)+....+\left(55-15\right)\left(55+15\right)\)
\(A=40.130+40.110+40.90+40.70\)
\(A=40.\left(130+110+90+70\right)=40.400=16000\)
b) \(B=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2011-2012\right)\left(2011+2012\right)\)
\(B=-3-7-11-...-4023\)
\(B=-\left(3+7+11+...+4023\right)\)
\(B=-\dfrac{\left(3+4023\right)\left[\dfrac{\left(4023-3\right)}{4}+1\right]}{2}=2025078\)
Ta cong them 1 vao moi phan thuc:
\(\frac{X+29}{1971}+1+\frac{X+27}{1973}+1+\frac{X+25}{1975}+1=\frac{X+1971}{29}+1+\frac{X+1973}{27}+1+\frac{X+1975}{25}+1\)
\(\Leftrightarrow\frac{X+2000}{1971}+\frac{X+2000}{1973}+\frac{X+2000}{1975}=\frac{X+2000}{29}+\frac{X+2000}{27}+\frac{X+2000}{25}\)
\(\Leftrightarrow\frac{X+2000}{1971}+\frac{X+2000}{1973}+\frac{X+2000}{1975}-\frac{X+2000}{29}-\frac{X+2000}{27}-\frac{X+2000}{25}=0\)
\(\Leftrightarrow\left(X+2000\right)\left(\frac{1}{1971}+\frac{1}{1973}+\frac{1}{1975}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}\right)=0\)
Vi \(\frac{1}{1971}+\frac{1}{1973}+\frac{1}{1975}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}\ne0\) nen \(X+2000=0\Leftrightarrow X=-2000\)
Vay \(X=-2000\)
Đề bài này khó hiểu vậy em ms hok lớp 6 thui ko biết đâu mà anh nào biết ghi ra để em xem với ạ