a: Đặt \(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\)
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}\)
...
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\)
Do đó: \(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}\)
=>\(A>\dfrac{100}{\sqrt{100}}=10\)
b: \(C=-\left|2x-6\right|-\left|3y+9\right|-18\)
\(\left|2x-6\right|>=0\forall x\)
\(\left|3y+9\right|>=0\forall y\)
Do đó: \(\left|2x-6\right|+\left|3y+9\right|>=0\forall x,y\)
=>\(-\left(\left|2x-6\right|+\left|3y+9\right|\right)< =0\forall x,y\)
=>\(-\left(\left|2x-6\right|+\left|3y+9\right|\right)-18< =-18\forall x,y\)
=>\(C< =-18\forall x,y\)
Dấu '=' xảy ra khi 2x-6=0 và 3y+9=0
=>x=3 và y=-3
