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A= ( 1/10-1) + ( 1/11 - 1 ) +...+ ( 1/100-1)
= 9/10 + 10/11 +...+ 99/100
= 9/100
^_^ ( have a good day)
\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(\Rightarrow B=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-98}{99}.\frac{-99}{100}\)
\(\Rightarrow B=\frac{-\left(1.2.3...98.99\right)}{2.3.4...99.100}\)
\(\Rightarrow B=\frac{-1}{100}\)
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{100}-1\right)\)
\(-A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{100}\right)\)
\(-A=\frac{9}{10}\cdot\frac{10}{11}\cdot\frac{11}{12}\cdot...\cdot\frac{99}{100}\)
\(-a=\frac{9}{100}\)
\(A=-\frac{9}{100}\)
Bài 1.
Ta có: \(\frac{a}{b}+\frac{-a}{b+1}=\frac{a}{b}-\frac{a}{b+1}=a\left(\frac{1}{b}-\frac{1}{b+1}\right)=a\left(\frac{b+1-b}{b\left(b+1\right)}\right)=a\left(\frac{1}{b\left(b+1\right)}\right)=\frac{a}{b\left(b+1\right)}\)
=> A là đáp án đúng
Bài 2. Ta có:
B = 4x - 4y + 5xy
B= 4x - 4y + 4xy + xy
B = 4(x - y + xy) + xy
B = 4.(5/12 - 1/3) - 1/3
B = 4.1/12 - 1/3
B = 1/3 - 1/3 = 0
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
\(\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{99}-1\right)\left(\frac{1}{100}-1\right)\)
\(=-\frac{9}{10}.\frac{-10}{11}.\frac{-11}{12}...\frac{-99}{100}\)
\(=-\frac{9}{100}\)