a) Vì a chia 13 dư 2 \(\Rightarrow\) a² chia 13 dư 4

b chia 13 dư 3 \(\Rightarrow\) b² chia 13 dư 9

\(\Rightarrow\) a² + b² chia hết cho 13

b) 10a² + 5b² + 12ab + 4a - 6b + 13

= ( 9a² + 12ab + 4b² ) + ( a² + 4a +4 ) + ( b² -6b + 9)

= (3a + 2b)² + (a + 2)² + (b - 3)²

Do (3a + 2b)² \(\overset{>}{-}\) 0

(a+ 2)² \(\overset{>}{-}\) 0

(b- 3)² \(\overset{>}{-}\) 0

\(\Rightarrow\) (3a + 2b)² + (a+ 2)² + (b- 3)² \(\overset{>}{-}\) 0

\(9a^2+12ab+4b^2+a^2+4a+4+b^2-6b+9=0\)

\(\Leftrightarrow\left(3a+2b\right)^2+\left(a+2\right)^2+\left(b-3\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}3a+2b=0\\a+2=0\\b-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-2\\b=3\end{matrix}\right.\)

\(M=\left(-4+3\right)^{2020}=1\)

\(1.CMR:\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)=1+\frac{b}{a}+\frac{a}{b}+1=\frac{a}{b}+\frac{b}{a}+2\)

Áp dụng BĐT AM-GM ta có:

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)

\(\Rightarrow\frac{a}{b}+\frac{b}{a}+2\ge2+2=4\)

Dấu '' = '' xảy ra khi \(a=b\)

\(2.\\ a.CMR:a^2+2b^2+c^2-2ab-2bc\ge0\forall a,b,c\)

\(a^2+2b^2+c^2-2ab-2bc=a^2-2ab+b^2+c^2-2bc+b^2=\left(a-b\right)^2+\left(b-c\right)^2\ge0\forall a,b,c\)

Dấu '' = '' xảy ra khi \(a=b=c\)

\(b.CMR:a^2+b^2-4a+6b+13\ge0\forall a,b\)

\(a^2+b^2-4a+6b+13=\left(a^2-4a+4\right)+\left(b^2+6b+9\right)=\left(a-2\right)^2+\left(b+9\right)^2\ge0\forall a,b\)

Dấu '' = '' xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=-9\end{matrix}\right.\)

câu 1 :a²+ab+ b²/4 +3b²/4=(a+b/2)² +3b²/2 tong 2 binh phương luôn >=0 dau bang khi ca hai số đó bằng 0. a=0 và b=0

câu 2: a^2-ab+ b²/4 +3b²/4=(a-b/2)² +3b²/2 .a=0 và b=0