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Phân tích phân số \(\dfrac{30}{43}\) ta có:
\(\dfrac{30}{43}=\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}\)
\(=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
Vậy: \(\left\{{}\begin{matrix}a=1\\b=2\\c=3\\d=4\end{matrix}\right.\)
\(\dfrac{30}{43}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\Leftrightarrow\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\\ \Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\\\Leftrightarrow\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}=\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\\d=4\end{matrix}\right.\)
Vậy............
Có \(\dfrac{30}{43}=\dfrac{1}{\dfrac{43}{30}}=\dfrac{1}{1+\dfrac{13}{30}}=\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{4}{13}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{13}{4}}}}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{4}}}}\)
Vậy a=1; b=2 ; c=3 ; d=4
ta thấy : \(\dfrac{a}{b}\) = \(\dfrac{1}{\dfrac{b}{a}}\)
\(\Rightarrow\) \(\dfrac{30}{43}\) = \(\dfrac{1}{\dfrac{43}{30}}\)
= \(\dfrac{1}{1+\dfrac{13}{30}}\)
= \(\dfrac{1}{1+\dfrac{1}{\dfrac{30}{13}}}\)
= \(\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{15}}}\)
= \(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{15}{2}}}}\)
=\(\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{7+\dfrac{1}{2}}}}\)
Vậy a = 1; b = 2; c = 7; d = 4
a: 51/56=1-5/56
61/66=1-5/66
mà -5/56<-5/66
nên 51/56<61/66
b: 41/43<1<172/165
c: \(\dfrac{101}{506}>0>-\dfrac{707}{3534}\)
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
a: 7/30=21/90
8/45=16/90
11/90=11/90
b: -4/5=-168/210
1/6=35/210
-9/7=-270/210
c: -7/24=-21/72
11/12=66/72
-23/36=-46/72
d: 17/30=85/150
-22/75=-44/150
5=750/150