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Ta có:
A=9999931999−5555571997
A=9999931998.999993−5555571996.555557
A=(9999932)999.999993 − (5555572)998.555557
A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)
A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)
A= \(\overline{\left(...0\right)}\)
Vì A có tận cùng là 0 nên \(A⋮5\)
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\dfrac{a-x}{a}=\dfrac{b-y}{b}\)
\(\Rightarrow1-\dfrac{x}{a}=1-\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{a}{b}\)
Có \(\dfrac{a}{b}=\dfrac{c}{d}< =>ad=bc\)
Xét \(\dfrac{a+c}{b+d}-\dfrac{a-c}{b-d}\)
= \(\dfrac{\left(a+c\right)\left(b-d\right)-\left(b+d\right)\left(a-c\right)}{\left(b+d\right)\left(b-d\right)}\)
= \(\dfrac{ab-ad+bc-cd-ab+bc-da+cd}{\left(b+d\right)\left(b-d\right)}\)
= 0
<=> \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Chọn đẳng thức \(\dfrac{d}{b}=\dfrac{c}{a}\) nhé bạn
sorry đăng nhầm,cái này mk hỏi có bn trả lời rồi
A=20/1.21+20/2.22+...+20/80.100
=1-1/21+1/2-1/22+...+1/80-1/100
=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)
80B=80/1.81+80/2.82+...+8/20.100
=1-1/81+1/2-1/82+...+1/20-1/100
=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)
=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)
=>20A=80B
=>A=4B
a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)
\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)
\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)
\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)
\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022