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Tách 1+2-3-4+5-6-8+...+97-98-99-100 với 101
A= 1 + 2 - 3 -4 + 5 + 6 -7 -8 + ... +97 +98 -99 -100 ( có: ( 100 - 1 ) : 1 + 1 = 100 )
A= ( 1 +2 - 3 - 4 ) + ( 5 + 6 - 7 -8 ) + ... ( 97 + 98 - 99 +100 ) ( có 100 : 4 = 25 cặp )
A= - 4 + -4 + -4 + ... + -4 ( có 25 số hạng )
A= ( -4 ) . 25
A= -100 + 101
A=1
học tốt
S=1/3+ - 1/4+ 1/5 + - 1/6 + 1/7 +1/6 + 1/-5 +1/4+1/-3
=(1/3+1/-3)+(-1/4+1/4)+(1/5+1/-5)+(-1/6+1/6)+1/7
=0+1/7
=1/7
25/31+ - 3/17+6/31+5/9+ - 14/17+ - 1/12
=(25/31+6/31)+(-3/17+-14/17)+(5/9+-1/12)
=1+(-1)+.....[tự tính 5/9+-1/12]
=0+........[kết quả trên]
=.........
1b) Ta có: \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{101}{100}=\frac{3.4.5....101}{2.3.4....100}=\frac{101}{2}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
\(P=4^{25}\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}\right)\cdot...\cdot\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{101}\right)\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{101}-\dfrac{1}{3}\right)\div2+1\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\dfrac{1}{50}\right)\)
\(P=4^{25}\cdot\dfrac{1}{100}\)