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Lời giải:
Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)
\(=99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)
\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)
$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$
Ta có đpcm.
Giải:
Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu
Ta có:
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\)
Chúc bạn học tốt!
Đặt A = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
3A = 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
4A = ( 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\) ) + ( \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) )
= 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
Đặt B = 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}\)
3B = 3 - 1 + \(\dfrac{1}{3}-\dfrac{1}{3^2}\) + ... - \(\dfrac{1}{3^{98}}\)
4B = ( 3 - 1 + \(\dfrac{1}{3}-\dfrac{1}{3^2}\) + ... - \(\dfrac{1}{3^{98}}\) ) + ( 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}\) )
= 3 - \(\dfrac{1}{3^{99}}\)
B = \(\dfrac{3}{4}-\dfrac{1}{3^{99}\cdot4}\)
⇒ 4A = \(\dfrac{3}{4}-\dfrac{1}{3^{99}\cdot4}\) - \(\dfrac{100}{3^{100}}\)
A = \(\dfrac{3}{16}-\dfrac{1}{3^{99}\cdot4^2}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
Vậy A < \(\dfrac{3}{16}\)
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Lời giải:
Đặt $P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2001}{3^{2001}}$
$3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2001}{3^{2000}}$
$3P-P=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}-\frac{2001}{3^{2001}}$
$2P+\frac{2001}{3^{2001}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2000}}$
$3(2P+\frac{2001}{3^{2001}})=3+1+\frac{1}{3}+...+\frac{1}{3^{1999}}$
$3(2P+\frac{2001}{3^{2001}})- (2P+\frac{2001}{3^{2001}})=3-\frac{1}{3^{2000}}$
$2(2P+\frac{2001}{3^{2001}}) =3-\frac{1}{3^{2000}}$
$P=\frac{1}{4}(3-\frac{4005}{3^{2001}})< \frac{3}{4}< \frac{4}{5}$