Tìm các số tự nhiên \(a,b\) biết: \(\dfrac{1719}{3976}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{a+\dfrac{1}{b}}}}}\)
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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
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\left(a,b\ne-1\right)\\ \Rightarrow2\left(a+b+2\right)=\left(a+1\right)\left(b+1\right)\\ \Rightarrow2a+2b+4=ab+a+b+1\\ \Rightarrow a+b-ab+3=0\\ \Rightarrow\left(b-1\right)-a\left(b-1\right)=-4\\ \Rightarrow\left(a-1\right)\left(b-1\right)=4=1\cdot4=2\cdot2\)
\(a-1\) | 1 | 4 | 2 |
\(b-1\) | 4 | 1 | 2 |
\(a\) | 2 | 5 | 3 |
\(b\) | 5 | 2 | 3 |
Vậy \(\left(a;b\right)=\left(2;5\right);\left(5;2\right);\left(3;3\right)\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\Leftrightarrow\dfrac{2\left(a+1\right)+2\left(b+1\right)-\left(a+1\right)\left(b+1\right)}{2\left(a+b\right)\left(b+1\right)}=0\)
\(\Leftrightarrow a+b-ab+3=0\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=-4\Leftrightarrow\left(a-1\right)\left(1-b\right)=-4\)
Do \(a,b\in N\) nên ta có bảng sau:
a-1 | -1 | 1 | -4 | 4 | -2 | 2 |
1-b | 4 | -4 | 1 | -1 | 2 | -2 |
a | 0 | 2 | -3(loại) | 5 | -1(loại) | 3 |
b | -3(loại) | 5 | 0 | 2 | -1(loại) | 3 |
Vậy \(\left(a;b\right)\in\left\{\left(2;5\right);\left(5;2\right);\left(3;3\right)\right\}\)
b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)
=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)
=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101
=>1-1/(n+1)=100/101
=>1/(n+1)=1/101
=>n+1=101
=>n=100
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
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: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
1.
=2/5 x 12/3 + 2/5 x 15/3 + 2/5 x 1
= 2/5 x (12/3 + 15/3 + 1)
=2/5 x 1
=2/5
2.a=1;2
Lời giải:
\(\frac{1719}{3976}=\frac{1}{2+\frac{538}{1719}}=\frac{1}{2+\frac{1}{3+\frac{105}{538}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{13}{105}}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{8+\frac{1}{13}}}}}\)
$\Rightarrow a=8; b=13$
\(\dfrac{1719}{3976}=\dfrac{1}{\dfrac{3976}{1719}}=\dfrac{1}{2+\dfrac{538}{1719}}=\dfrac{1}{2+\dfrac{1}{\dfrac{1719}{538}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{105}{538}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{\dfrac{538}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{13}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{\dfrac{105}{13}}}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{8+\dfrac{1}{13}}}}}\)