Tìm các số tự nhiên a, b, c sao cho :
\(a,\dfrac{a}{3}+\dfrac{b}{4}=\dfrac{a+b}{3+4}\)
\(b,\dfrac{52}{9}=5+\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{c}}}\)
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Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
Lời giải:
\(\frac{1719}{3976}=\frac{1}{2+\frac{538}{1719}}=\frac{1}{2+\frac{1}{3+\frac{105}{538}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{13}{105}}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{8+\frac{1}{13}}}}}\)
$\Rightarrow a=8; b=13$
\(\dfrac{1719}{3976}=\dfrac{1}{\dfrac{3976}{1719}}=\dfrac{1}{2+\dfrac{538}{1719}}=\dfrac{1}{2+\dfrac{1}{\dfrac{1719}{538}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{105}{538}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{\dfrac{538}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{13}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{\dfrac{105}{13}}}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{8+\dfrac{1}{13}}}}}\)
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
a,a+1/4=2 3/4-1 1/2
a+1/2=5/4
a=5/4-1/2
a=3/4
b,a-7/4=13/4-7/9
a-7/4=89/36
a= 89/36+7/4
a=152/36
c,3/2-a=17/6-1/6
3/2-a=8/3
a= 3/2-8/3
a= -7/6
a: \(\Leftrightarrow-4< =x< =-3\)
hay \(x\in\varnothing\)
b: =>-9<x<=3
hay \(x\in\left\{0;1;2;3\right\}\)