We see: 

\(4=-2+6\)

\(16=4+6+6\)

So each number in the following arithmetic is less than 6 units than the after number. So \(m=4+6=10\)

\(n^2+7n+22=\left(n^2-2n+1\right)+9n+21=\left(n-1\right)^2+9n+21\)

\(=\left(n-1\right)^2-3+9\left(n+2\right)\)

\(n^2+7n+22⋮9\Leftrightarrow\left(n-1\right)^2-3⋮9\)

Or \(\left(n-1\right)^2=9k+12\)\(=3\left(3k+2\right)\)

Because there are no square number in the form 3k+2

So \(n^2+7n+22⋮̸9\)

The integer which is multiplies of neither 2 nor 5 is ending by 0

So they're: 10,20,30,40,50,60,70,80,90,100

Their sum is: 10+20+...+100 = 550

Oh, right Alone

We have:

\(a\ge3\) and \(b\ge3\)

So \(a+b\ge6\)

If a + b = 6 then a = 3; b = 3

\(\Rightarrow a^2+b^2=3^2+3^2=18< 25\)

So \(a+b>6\)

Or \(a+b\ge7\)

We have:

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}\)

Apply the same sequence properties, we have:

\(\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}=\dfrac{xy+yz+xz}{24+30+80}=\dfrac{1206}{134}=9\)

So \(x^2=9.64=576\)\(\Rightarrow x=24\)

\(\dfrac{xy}{24}=9\Rightarrow xy=24.9=216\Rightarrow y=9\)

\(\dfrac{yz}{30}=9\Rightarrow yz=270\Rightarrow z=30\)

So \(\left(x;y;z\right)=\left(24;9;30\right)\)

We see:

\(\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\)

\(\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\)

\(\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\)( Because a,b,c > 0 )

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a+b+c}{a+b+c}=1\left(1\right)\)

Else: We have: \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(a< b,b\ne0;n>0\right)\)

So \(\left\{{}\begin{matrix}\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\\\dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\\\dfrac{c}{a+c}< \dfrac{b+c}{a+b+c}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c+a+b+b+c}{a+b+c}=2\left(2\right)\)

From (1) and (2) we have:

\(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

No pairs satisfy

According to the title, we have:

99<a<110 and a is a prime number

So a=103

b is the greatest 1-digit number so b = 9

a+b=103+9=112

We have:

\(xy=1\Leftrightarrow y=\dfrac{1}{x}\)

So \(\left|x+y\right|=\left|\dfrac{1}{x}+x\right|\)

Other way:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\left(a,b\ne0\right)\)

So \(x+\dfrac{1}{x}\ge2\)

And \(\left|x+\dfrac{1}{x}\right|\ge2\)

Or \(\left|x+y\right|\ge2\)

\(\left|x+y\right|_{Min}=2\Leftrightarrow x=y=1\)

We have:

\(A=\left|x-3\right|+\left|x-7\right|\)

\(=\left|x-3\right|+\left|7-x\right|\)

\(\ge\left|x-3+7-x\right|=4\)

Equal sign occurs \(\Leftrightarrow\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\7-x< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x>7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

So \(A_{Min}=4\)\(\Leftrightarrow3\le x\le7\)  :)

We see:

\(n!=1.2.3....n\)

So \(\left(n!\right)^2=\left(1.2.3.....n\right)^2\)

Because n>2 so n! > 1.2.n 

So \(\left(n!\right)^2>\left(1.2.n\right)^2=4.n^2>n^2\)

Suppose: If all 9 numbers is smaller or equal than 1

We see: 

The sum of them is small or equal than 9

So at least one of them must be bigger than 1 to have the sum 10 :)

She has the same number of apples and oranges

\(\Rightarrow x=4y\)

Or \(\dfrac{x}{y}=\dfrac{4}{1}\)

We have:

\(8\sqrt{3}=\sqrt{8^2.3}=\sqrt{64.3}=\sqrt{192}\)

\(5\sqrt{7}=\sqrt{5^2.7}=\sqrt{25.7}=\sqrt{175}\)

Because 192>175 so \(\sqrt{192}>\sqrt{175}\)

or \(8\sqrt{3}>5\sqrt{7}\) 

After using strawberries to make cake and make juice, Anna's strawberries left is: 

708-12-20=676(strawberries)

Anna is only has:

676:708\(\approx0,95=95\%\)

We have:

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)

Other way:

\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)

Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)

\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)

So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)

or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)

From (*) and (**) we have  \(a+b\le2\)

Your ex is so hard to do :)