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\(a,\dfrac{1}{2}-\dfrac{1}{3}-\left(-\dfrac{5}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{5}{4}=\dfrac{1\times6-1\times4+5\times3}{12}=\dfrac{6-4+15}{12}=\dfrac{17}{12}\\ b,\dfrac{5}{4}-\dfrac{1}{2}-\dfrac{7}{8}=\dfrac{5\times2-1\times4-7}{8}=\dfrac{10-4-7}{8}=-\dfrac{1}{8}\\ c,\dfrac{1}{5}-\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{1\times2-1\times5+9}{10}=\dfrac{2-5+9}{10}=\dfrac{6}{10}=\dfrac{3}{5}\\ d,\dfrac{5}{4}-\dfrac{1}{3}+\dfrac{7}{6}=\dfrac{5\times3-1\times4+7\times2}{12}=\dfrac{15-4+14}{12}=\dfrac{25}{12}\)
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A = 100 + 98 + 96 + ... + 2 - 97 - 95 - ... - 1
A = 100 + (98 - 97) + (96 - 95) + ... + (2 - 1)
A = 100 + 1 + 1 + ... + 1 (49 số 1)
A = 100 + 49 = 149
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Đặt A = 5 × 6 + 6 × 7 + 7 × 8 + ... + 99 × 100
3A = 5 × 6 × (7 - 4) + 6 × 7 × (8 - 5) + 7 × 8 × (9 - 6) + ... + 99 × 100 × (101 - 98)
3A = 5 × 6 × 7 - 4 × 5 × 6 + 6 × 7 × 8 + 7 × 8 × 9 - 6 × 7 × 8 + .... + 99 × 100 × 101 - 98 × 99 × 100
3A = 99 × 100 × 101 - 4 × 5 × 6
3A = 3 x (33 × 100 × 101 - 4 × 5 × 2)
A = 33 × 100 × 101 - 4 × 5 × 2
A = 333300 - 40
A = 333260
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Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{98^2}\)
Ta có : \(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{97.98}\)
\(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{97}-\frac{1}{98}\)
\(A< \frac{1}{2}-\frac{1}{98}\)
\(A< \frac{1}{2}\)
GOOD LUCK !!!
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\(1-2+3-4+5-6+...+97-98+99-100\)
\(=\left(1+3+5+...+97+99\right)-\left(2+4+6+...+98+100\right)\)
\(=\left[\left(99+1\right)\times\frac{\left(99-1\right)\div2+1}{2}\right]-\left[\left(98+2\right)\times\frac{\left(98-2\right)\div2+1}{2}\right]\)
\(=\left(100\times25\right)-\left(100\times\frac{49}{2}\right)\)
\(=2500-2450\)
\(=50\)
=(3-1)+(7-5)+...+(99-97) (có 50 cặp)
=2+2+..+2
=2*50
=100
-1 + 3- 5 + 7-...-97 + 99 ( có 50 số hạng )
= (3 - 1)+(7 - 5)+ ... +(99 - 97) ( có 25 nhóm )
= 2 + 2 + ... + 2 (có 25 số hạng 2)
= 25.2
=50