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Đây là cuộc thi nhé. cần sự công bằng. Mong em không tái phạm lần sau. Bạn sẽ bị khóa nick hoặc trừ 5000 điểm nhé!
BQT thân gửi em!
__BQT Lớp 6/7 Hỏi Đáp__
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\(A=\left(1985\cdot1987-1\right):\left(1980+1985\cdot1986\right)\)
\(A=3944194\div3944190\)
ko chia hết nên sẽ bằng 1,4 lớn hơn 1
\(\Rightarrow A>1\)
1985x1987-1/1980+1985x1986=1985x1986+1985-1/1980+1985x1986
=1985x1986+1984/1980+1985x1986.Vì 1985x1986+1984>1980+1985x1986
suy ra 1985x1987-1/1980+1985x1986>1
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\(\frac{1985\times1987-1}{1984+1985\times1986}\)
\(=\frac{1985\times\left(1986+1\right)-1}{1984+1985\times1986}\)
\(=\frac{1985\times1986+1985-1}{1984+1985\times1986}\)
\(=\frac{1985\times1986+1984}{1984+1985\times1986}\)
\(=1\)
----------HOK TỐT-----------
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Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(A=\frac{19^{18}+1}{19^{19}+1}< \frac{19^{18}+1+18}{19^{19}+1+18}=\frac{19^{18}+19}{19^{19}+19}=\frac{19\left(19^{17}+1\right)}{19\left(19^{18}+1\right)}=\frac{19^{17}+1}{19^{18}+1}=B\)
\(\Rightarrow\)\(A< B\) ( đpcm )
Vậy \(A< B\)
Chúc bạn học tốt ~
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1984 x 1985 + 1986 x 20 + 1985
= 1984 x 1985 + ( 1985 + 1 ) x 20 + 1985
= 1984 x 1985 + 1985 x 20 + 20 + 1985
= 1985 x ( 1984 + 1 + 20 ) + 20
= 1985 x 2005 + 20
= 3979925 + 20
= 3979945
1985 x 2000 - 1984 x 1985
= 1985 x ( 2000 - 1984 )
= 1985 x 16
= 31760
Hok tốt !
Ta có: \(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)
\(=\frac{124}{1984}\left(\frac{1984}{1.1985}+\frac{1984}{2.1986}+\frac{1984}{3.1987}+...+\frac{1984}{16.2000}\right)\)
\(=\frac{1}{16}\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\right]\)
\(B=\frac{1}{1.17}+\frac{1}{2.19}+...+\frac{1}{1984.2000}\)
\(=\frac{1}{16}\left(\frac{16}{1.17}+\frac{16}{2.18}+...+\frac{16}{1984.2000}\right)\)
\(=\frac{1}{16}\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)\right]-\left[\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right]\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
\(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)
Vậy A = B
dễ tự nghĩ