Btinh gia tri bieu thuc B=(1/2^2 -1) . ( 1/3^2-1) .(1/4^2 -1) . ... .(1/98 ^2 -1) . (1/99^2 -1)
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\(D=\left(-\dfrac{1}{3}\right)^1+\left(-\dfrac{1}{3}\right)^2+...+\left(-\dfrac{1}{3}\right)^{98}+\left(\dfrac{-1}{3}\right)^{99}\)
\(\Leftrightarrow\left(-\dfrac{1}{3}D\right)=\left(-\dfrac{1}{3}\right)^2+...+\left(-\dfrac{1}{3}\right)^{99}+\left(-\dfrac{1}{3}\right)^{100}\)
\(\Leftrightarrow D\cdot\dfrac{-4}{3}=\dfrac{1^{100}}{3^{100}}-\left(-\dfrac{1}{3}\right)=\dfrac{1}{3^{100}}+\dfrac{1}{3}=\dfrac{1+3^{99}}{3^{100}}\)
\(\Leftrightarrow D=\dfrac{3^{99}+1}{3^{100}}:\dfrac{-4}{3}=\dfrac{3^{99}+1}{-4\cdot3^{99}}\)

Ta có : \(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).......\left(1+\frac{1}{3}\right)\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.\frac{99}{98}......\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)
\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right).....\left(1+\frac{1}{3}\right).\left(1+\frac{1}{2}\right)\)
\(=\frac{101}{100}.\frac{100}{99}.....\frac{4}{3}.\frac{3}{2}=\frac{101}{2}\)


Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
B = ( 1/2^2 - 1) . ( 1/3^2 - 1) . ( 1/4^2 - 1) ... ( 1/98^2 - 1) . ( 1/99^2 - 1)
B = -3/2^2 . ( -8/3^2) . ( -15/4^2) .... ( -9603/98^2) . ( -9800/99^2)
Tích B gồm ( 99 - 2) : 1 + 1 = 98 ( số hạng), mỗi số hạng đều mang dấu âm nên khi tính ra kết quả tích B mang dấu dương
=> B = 3/2^2 . 8/3^2 . 15/4^2 ... 9603/98^2 . 9800/99^2
B = 1.3/2.2 . 2.4/3.3 . 3.5/4.4 ... 97.99/98.98 . 98.100/99.99
B = 1.2.3...97.98/2.3.4...98.99 . 3.4.5...99.100/2.3.4...98.99
B = 1/99 . 100/2
B = 1/99 . 50 = 50/99