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a chia 5 dư 2=>a=5q+2
b chia 5 dư 3=>b=5k+3
=>ab=(5q+2)(5k+3)=(5q+2)5+(5q+2)3
=(5q+2)5+5.3q+6 chia 5 dư 1
Vậy ab chia 5 dư 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 2a+2b=10b+a suy ra a=8b
Mà a+b=căn ab
=> a2+2ab+b2=10a+b
=>(8b)2+2(8b)b+b2=10(8b)+b
=>81b2=81b
Lại có b>0 =>b=1 =>a=8
Vậy a-b=7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a-b=8\Rightarrow\left(a-b\right)^2=8^2=64\)
\(\Rightarrow a^2-2ab+b^2=64\)
\(\Rightarrow a^2+b^2-2.10=64\)
\(\Rightarrow a^2+b^2=84\)
\(\Rightarrow a^2+2ab+b^2=84+2.10=84+20=104\)
\(\Rightarrow\left(a+b\right)^2=104\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Ta có:}\)
\(a^{12}+b^{12}-\left(a^{11}+b^{11}\right)\left(a+b\right)+\left(a^{10}+b^{10}\right)ab=0\)
\(\Rightarrow\left(a^{12}+b^{12}\right)\left(ab-a-b+1\right)=0\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)
\(+,a=1\Rightarrow b^{10}=b^{11}=b^{12}\Rightarrow b=1\left(\text{vì b dương}\right)\)
\(+,b=1\Rightarrow a^{10}=a^{11}=a^{12}\Rightarrow a=1\left(\text{vì a dương}\right)\)
\(\text{nên: a=b=1}\)
\(\text{Vậy: P=2011a-2012b=2011-2012=-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a-b=3\Rightarrow\left(a-b\right)^2=9\Rightarrow a^2-2ab+b^2=9\Rightarrow a^2+2ab+b^2-4ab=9\)
\(\Rightarrow\left(a+b\right)^2-4ab=9\)
Thay (a + b)2 = 10 vào ta có:
\(10-4ab=9\Rightarrow4ab=1\Rightarrow ab=\frac{1}{4}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b=17
=>(a+b)^2=289
Mà (a+b)^2=a^2+2ab+b^2; ab=72
=>a^2+b^2=289-2x72=145
(a+b)2 =102 \(\Leftrightarrow a^2+2ab+b^2=100\Leftrightarrow a^2-2ab+b^2+4ab=100\)
\(\Leftrightarrow\left(a-b\right)^2=100-4ab\Leftrightarrow\left(a-b\right)^2=144\)