Tìm các số tự nhiên \(a,b,c,d\) biết rằng: \(\dfrac{25112012}{11}=a-\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{d}}}\)
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\(\dfrac{1}{a+1}+\dfrac{1}{b+1}=\dfrac{1}{2}\left(a,b\ne-1\right)\\ \Rightarrow2\left(a+b+2\right)=\left(a+1\right)\left(b+1\right)\\ \Rightarrow2a+2b+4=ab+a+b+1\\ \Rightarrow a+b-ab+3=0\\ \Rightarrow\left(b-1\right)-a\left(b-1\right)=-4\\ \Rightarrow\left(a-1\right)\left(b-1\right)=4=1\cdot4=2\cdot2\)
\(a-1\) | 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 \(\left(a;b\right)\in\left\{\left(2;5\right);\left(5;2\right);\left(3;3\right)\right\}\)
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
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
\(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
<=>b(c+d)(d+a)+d(a+b)(b+c)=0 (vì c≠a)
<=>abc-acd+bd2-b2d=0
<=> (b-d)(ac-bd)=0 <=> ac - bd =0 (vì b≠d) <=> ac = bd
Vậy abcd =(ac)(bd)=(ac)2
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
a) Ta có: \(\dfrac{3+x}{7+y}=\dfrac{3}{7}\)
\(\Leftrightarrow\dfrac{x+3}{3}=\dfrac{y+7}{7}\)
mà x+y=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+3}{3}=\dfrac{y+7}{7}=\dfrac{x+y+3+7}{3+7}=\dfrac{20+10}{10}=3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x+3}{10}=3\\\dfrac{y+7}{7}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=30\\y+7=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=14\end{matrix}\right.\)
Vậy: x=27; y=14
Lời giải:
\(\frac{1719}{3976}=\frac{1}{2+\frac{538}{1719}}=\frac{1}{2+\frac{1}{3+\frac{105}{538}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{13}{105}}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{8+\frac{1}{13}}}}}\)
$\Rightarrow a=8; b=13$
\(\dfrac{1719}{3976}=\dfrac{1}{\dfrac{3976}{1719}}=\dfrac{1}{2+\dfrac{538}{1719}}=\dfrac{1}{2+\dfrac{1}{\dfrac{1719}{538}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{105}{538}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{\dfrac{538}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{13}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{\dfrac{105}{13}}}}}\)
\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{8+\dfrac{1}{13}}}}}\)
\(\dfrac{25112012}{11}=2282911-\dfrac{9}{11}\Rightarrow a=2282911\)
\(\dfrac{9}{11}=\dfrac{1}{\dfrac{11}{9}}=\dfrac{1}{1+\dfrac{2}{9}}=\dfrac{1}{1+\dfrac{1}{\dfrac{9}{2}}}=\dfrac{1}{1+\dfrac{1}{4+\dfrac{1}{2}}}\)
\(\Rightarrow b=1;c=4;d=2\)