Tính giá trị:
\(P=\sqrt{1+1999^2+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
Những câu hỏi liên quan
Tính A = \(\sqrt{1+1999^2+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
Đặt 2000 = a thì ta có
A = \(\sqrt{1+\left(a-1\right)^2+\frac{\left(a-1\right)^2}{a^2}}+\frac{a-1}{a}\)
\(=\sqrt{\frac{a^4-2a^3+3a^2-2a+1}{a^2}}+\frac{a-1}{a}\)
\(=\frac{a^2-a+1}{a}+\frac{a-1}{a}=a=2000\)
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Rút gọn A = \(\sqrt{1+1999^2+\frac{1999^2}{2000^2}}+\frac{1999}{2000}\)
tính:A=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2000\sqrt{1999}+1999\sqrt{2000}}\)
\(\frac{1}{n\sqrt{n+1}+\sqrt{n}\left(n+1\right)}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)
sau đó tách ra là ok
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A=\(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1999}{2}+\frac{1998}{3}+....+\frac{1}{2000}}\)
Các bạn giải dùm mình nha
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}}\)
\(=\frac{\left[\frac{2001}{1}+1\right]+\left[\frac{2001}{2}+1\right]+...+\left[\frac{2001}{2000}+1\right]+2001}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}\)
\(=\frac{2001\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}\right]}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}=2001\)
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tính: P=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+...+\frac{1}{1999}}\)
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Tính nhanh:\(\frac{1^2+1}{2^2+1}+\frac{1^2-1}{2^2-1}-\frac{2^2+1}{3^2+1}+\frac{2^2-1}{3^2-1}+...-\frac{1999^2+1}{2000^2+1}+\frac{1999^2-1}{2000^2-1}\)
Tính \(E=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}}\)
Gọi 1/4 số a là 0,25 . Ta có :
a . 3 - a . 0,25 = 147,07
a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )
a . 2,75 = 147,07
a = 147,07 : 2,75
a = 53,48
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Tính
A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+......+\frac{1}{1999}}\)
Ai nhanh và đúng mình tick cho
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)
\(=\frac{1}{2000}\)
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5 tháng 6 2015 lúc 10:14
tìm x, biết:
\(\frac{x-1}{2000}+\frac{x-2}{1999}+\frac{x-3}{1998}+...+\frac{x-1999}{2}=1999\)
\(\Leftrightarrow\frac{x-1}{2000}-1+\frac{x-2}{1999}-1+\frac{x-3}{1998}-1+....+\frac{x-1999}{2}-1=0\)
\(\Leftrightarrow\frac{x-2001}{2000}+\frac{x-2001}{1999}+\frac{x-2001}{1998}+....+\frac{x-2001}{2}=0\)
\(\Leftrightarrow\left(x-2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+...+\frac{1}{2}\right)=0\)
\(\Leftrightarrow x-2001=0\)
\(\Leftrightarrow x=2001\)
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