Tìm \(\overline{ab}\), biết:
\(\sqrt{\overline{ab}}=a+b\)
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Bài 3.
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=-\dfrac{1}{24}\left(1\right)\\c\left(a+b+c\right)=-\dfrac{1}{72}\left(2\right)\\b\left(a+b+c\right)=\dfrac{1}{16}\left(3\right)\end{matrix}\right.\)
Dễ thấy \(a,b,c\ne0\Rightarrow a+b+c\ne0\)
Chia (1) cho (2), ta được \(\dfrac{a}{c}=3\Rightarrow a=3c\left(4\right)\)
Chia (2) cho (3) ta được: \(\dfrac{c}{b}=-\dfrac{2}{9}\Rightarrow b=-\dfrac{9}{2}c\left(5\right)\).
Thay (4), (5) vào (2), ta được: \(-\dfrac{1}{2}c^2=-\dfrac{1}{72}\)
\(\Rightarrow c=\pm\dfrac{1}{6}\).
Với \(c=\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=\dfrac{1}{2}\\b=-\dfrac{9}{2}c=-\dfrac{3}{4}\end{matrix}\right.\)
Với \(c=-\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=-\dfrac{1}{2}\\b=-\dfrac{9}{2}c=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(\left(a;b;c\right)=\left\{\left(\dfrac{1}{2};-\dfrac{3}{4};\dfrac{1}{6}\right);\left(-\dfrac{1}{2};\dfrac{3}{4};-\dfrac{1}{6}\right)\right\}\)
Ta có :
\(\overline{a,b}.\overline{ab,a}=\overline{ab,ab}\)
\(\Leftrightarrow\)\(\left(\overline{a,b}.10\right)\left(\overline{ab,a}.10\right)=\overline{ab,ab}.100\)
\(\Leftrightarrow\)\(\overline{ab}.\overline{aba}=\overline{abab}\)
\(\Leftrightarrow\)\(\overline{ab}.\overline{aba}=\overline{ab}.\left(100+1\right)\)
\(\Leftrightarrow\)\(\overline{aba}=101\)
\(\Rightarrow\)\(a=1\)\(;\)\(b=0\)
Vậy \(a=1\) và \(b=0\)
Đáp án:
1352013520 hoặc 63504.63504.
Giải thích các bước giải:
¯¯¯¯¯¯¯¯¯¯¯¯¯abcde=2¯¯¯¯¯ab.¯¯¯¯¯¯¯¯cde⇒1000¯¯¯¯¯ab+¯¯¯¯¯¯¯¯cde=2¯¯¯¯¯ab.¯¯¯¯¯¯¯¯cde⇒1000¯¯¯¯¯ab=−¯¯¯¯¯¯¯¯cde+2¯¯¯¯¯ab.¯¯¯¯¯¯¯¯cde⇒1000¯¯¯¯¯ab=(2¯¯¯¯¯ab−1)¯¯¯¯¯¯¯¯cde(∗)⇒1000¯¯¯¯¯ab ⋮ 2¯¯¯¯¯ab−1�����¯=2��¯.���¯⇒1000��¯+���¯=2��¯.���¯⇒1000��¯=−���¯+2��¯.���¯⇒1000��¯=(2��¯−1)���¯(∗)⇒1000��¯ ⋮ 2��¯−1
Do (¯¯¯¯¯ab;2¯¯¯¯¯ab−1)=1(��¯;2��¯−1)=1
⇒1000 ⋮ 2¯¯¯¯¯ab−1⇒1000 ⋮ 2��¯−1
2¯¯¯¯¯ab−1≥19(¯¯¯¯¯ab2��¯−1≥19(��¯ nhỏ nhất là 10)10)
Ước dương của 10001000
Ư(1000)={1;2;4;5;8;10;20;25;40;50;100;125;200;250;500;1000}Ư(1000)={1;2;4;5;8;10;20;25;40;50;100;125;200;250;500;1000}
Do 2¯¯¯¯¯ab−12��¯−1 lẻ và 2¯¯¯¯¯ab−1≥192��¯−1≥19
⇒(2¯¯¯¯¯ab−1)∈{25;125}⊛2¯¯¯¯¯ab−1=25⇒2¯¯¯¯¯ab=26⇒¯¯¯¯¯ab=13(∗)⇒1000.13=(2.13−1)¯¯¯¯¯¯¯¯cde⇒13000=25¯¯¯¯¯¯¯¯cde⇒¯¯¯¯¯¯¯¯cde=520⊛2¯¯¯¯¯ab−1=125⇒2¯¯¯¯¯ab=126⇒¯¯¯¯¯ab=63(∗)⇒1000.63=(2.63−1)¯¯¯¯¯¯¯¯cde⇒63000=125¯¯¯¯¯¯¯¯cde⇒¯¯¯¯¯¯¯¯cde=504⇒(2��¯−1)∈{25;125}⊛2��¯−1=25⇒2��¯=26⇒��¯=13(∗)⇒1000.13=(2.13−1)���¯⇒13000=25���¯⇒���¯=520⊛2��¯−1=125⇒2��¯=126⇒��¯=63(∗)⇒1000.63=(2.63−1)���¯⇒63000=125���¯⇒���¯=504
Vậy số thoả mãn là 1352013520 hoặc 63504.
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}-\overline{bc}-\overline{ca}+\overline{ca}+\overline{ab}}{a+b-b-c+c+a}=\frac{2\overline{ab}}{2a}=10+\frac{b}{a}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}+\overline{bc}+\overline{ca}-\overline{ca}-\overline{ab}}{a+b+b+c-c-a}=\frac{2\overline{bc}}{2b}=10+\frac{c}{b}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{-\overline{ab}-\overline{bc}+\overline{bc}+\overline{ca}+\overline{ca}+\overline{ab}}{-a-b+b+c+c+a}=\frac{2\overline{ca}}{2c}=10+\frac{a}{c}\)
=> \(\frac{b}{a}=\frac{c}{b}=\frac{a}{c}\Rightarrow\frac{b+c+a}{a+b+c}=1\Rightarrow a=b=c\)
\(\sqrt{ab}=a+b\)
a^2+2ab+b^2=10a+b
a^2+2(b-5)a+b^2-b=0
a^2+2(b-5)a+(b-5)^2+9b-25=0
(a+(b-5)^2=25-9b
(a+(b-5)^2>=0\(\hept{\begin{cases}25-9b\ge0\Rightarrow b\le3\\25-9b=k^2\Rightarrow b=\left\{0,1\right\}\end{cases}}\)
\(b=0\Rightarrow\left(a-5\right)^2=25\Rightarrow\orbr{\begin{cases}a=0\left(loai\right)\\a=10\end{cases}}\)
\(b=1\Rightarrow\left(a-4\right)^2=16\Rightarrow\orbr{\begin{cases}a=0\left(loai\right)\\a=8\end{cases}}\)
Kết luận:
ab =100
ab=81