Call the bigger numbers a, the smaller number b

We have: a-b=6 and a+b=24

=> a-b+a+b=6+24

=>2a=30

=>a=30:2

=>a=15

=> 15-b=6

=>b=9

So the products of the two numbers are 15 and 9

\(0,0A=\sqrt{0.0049}=0.07\)

\(\Rightarrow A=7\)

Give: \(a+b+c\ge abc\).Prove that:

\(a^2+b^2+c^2\ge abc\)

Let a , b and c be positive real numbers such that \(a^2+b^2+c^2=12\) . Prove that \(\dfrac{a+b}{4+bc}+\dfrac{b+c}{4+ca}+\dfrac{c+a}{4+ab}\ge\dfrac{3}{2}\) .

1+2+3+...+10

=(1+10)*10:2

=11*5

=55

Prove: \(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ca+b^2}+\dfrac{c+a}{ab+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\left(a;b;c>0\right)\)

Solve the equation:

\(x.\dfrac{8-x}{x-1}\left(x-\dfrac{8-x}{x-1}\right)=15\)

\(C=\dfrac{3\left|x\right|+2}{4\left|x\right|-5}\)         \(\left(x\in Z\right)\)

a)Find Max,Min C

b) Find x when \(C\in N\)

Give: \(a+b+c=4\).Prove:

\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}>4\)

Solve these equations:

a) \(\dfrac{6}{\left(x+1\right)\left(x+2\right)}+\dfrac{8}{\left(x-1\right)\left(x+4\right)}=1\)

b)\(\dfrac{x^2-3x+5}{x^2-4x+5}+\dfrac{x^2-5x+5}{x^2-6x+5}=\dfrac{-1}{4}\)