a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

Áp dụng Côsi

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

Tương tự: \(\frac{bc}{a}+\frac{ca}{b}\ge2c;\frac{ca}{b}+\frac{ab}{c}\ge2a\)

\(\Rightarrow2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)=2\)

\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge1\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)

Vậy GTNN của A là 1

kho qua ai biet giai khong

a) \(aaaa:x=a\Rightarrow aaaa:a=x\Rightarrow x=1111\)

b) \(x\times a=a0a0a0\Rightarrow x=a0a0a0:a\Rightarrow x=101010\)

giup minh voi