\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=9\\a^4+2a^2b^2+b^4\ge25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2ab=9-\left(a^2+b^2\right)\\a^4+b^4+2a^2b^2\ge25\end{matrix}\right.\)

We have : a² + b² \(\ge\) 5

=> \(-\left(a^2+b^2\right)\le5\)

=> \(9-\left(a^2+b^2\right)\ge9-5=4\)

=> \(2ab\ge4\)

=> \(ab\ge2\)

<=> \(a^2b^2\ge4\)

<=> \(4a^2b^2\ge16\)

Plus \(4a^2b^2\ge16\) into \(a^4+b^4+2a^2b^2\ge25\)

=> \(a^4+b^4+6a^2b^2\ge41\)

=> Min = 41

That is my opinion :v 

Uầy... từ từ .... coi chừng gây war đó , Summer Clouds , nhờ cô :< 

Đồng ý với @WhySoSerious

That is my question ok : https://mathulike.com/questions/1902.html

Searching before asking , ok cheater :V

Apply Cauchy inequality , we have :

\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\dfrac{4}{5}\left(x+y\right)+\left(\dfrac{6x}{5}+\dfrac{30}{x}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)\)

\(P\ge\dfrac{4}{5}.10+2.\sqrt{\dfrac{6x}{5}.\dfrac{30}{x}}+2.\sqrt{\dfrac{y}{5}.\dfrac{5}{y}}=8+6+2=16\)

So \(Min_P=16\)

<=> x = y = 5

With the formula for calculating time is : \(t=0,45.\sqrt{d}\) \(\left\{{}\begin{matrix}t=time\\d=distance\end{matrix}\right.\)

So that , with the distance of 200m , Time to rock falls is :

\(0,45.\sqrt{200}=0,45.10\sqrt{2}=\dfrac{9\sqrt{2}}{2}\approx6,36\) (seconds)

\(1-5\cdot2+10\cdot2^2-10\cdot2^3+5\cdot2^4-2^5\)

\(=1-5\cdot2+5\cdot2^3-5\cdot2^4+5\cdot2^4-2^5\)

\(=1-5\cdot2+5\cdot2^3-2^5\)

\(=1-5\cdot\left(2-2^3\right)-2^5\)

\(=1-5.\left(-6\right)-32\)

\(=1+30-32=-1\)

\(\dfrac{64^8}{32^{-3}}=\dfrac{\left(2^6\right)^8}{\dfrac{1}{32^3}}=2k\)

\(2^{48}.\left(2^5\right)^3=2k\)

\(2^{48}.2^{15}=2k\)

\(2^{48+15}=2^{63}=2^k\)

=> k = 63 

Ohh.....there are some problems here :v 

At tenth round, change to : P/s: We do l-i-k-e this to responsive the question:

The question says that : one of whom needs to row to cross a river

That means , at least 1 person crosses the river.

In first trip , the boat holds three people and takes them to the right bank of the river side. Then, the boat turned to the left side of the river with two people sat on the boat.

Similar with second trip, third trip ........

At last , the boat takes three peoples to the right bank and end the trip.

So, there are 7 trips go and come back .

That mean, the maximum of the distance the boat must travelto transport 9 people is :

(7 . 2) . 20 = 14.20 = 280 (yards) 

P/s: We do that to responsive the question:

1. The maximum distance the boat must travel , that means we have to find how many times the boat goes most

2. At this : P/s: There must be at least 2 people on boat to run it  :) 

The question says that : one of whom needs to row to cross a river

That means , at least 1 person crosses the river.

In first trip , the boat holds three people and takes them to the right bank of the river side. Then, the boat turned to the left side of the river with two people sat on the boat.

Similar with second trip, third trip ........

At last , the boat takes three peoples to the right bank and end the trip.

So, there are 7 trips go and come back .

That mean, the maximum of the distance the boat must travelto transport 9 people is :

(7 . 2) . 20 = 14.20 = 280 (yards) 

P/s: We do this to responsive the question:

1. The maximum distance the boat must travel , that means we have to find how many times the boat goes most

2. At this : P/s: There must be at least 2 people on boat to run it  :) 

Change : 18 seconds = \(\dfrac{18}{60}\left(min\right)=0,3\left(min\right)\)

In 1 minute , the wheel travel : \(15:\dfrac{3}{10}=\dfrac{15.10}{3}=5.10=50\left(feet\right)\)

The perimeter of the wheel is: \(\dfrac{50}{10}=5\left(feet\right)\)

The diameter of the wheel is : \(\dfrac{5}{pi}=\dfrac{5}{3,14}=1,59....\approx1,6\left(feet\right)\)

With x^2x = 1

\(\Rightarrow\left[{}\begin{matrix}x=1\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

But the question asked how many non-zero values of x 

So that : x = 1 

\(45+\sqrt{c}=49\)

<=> \(\sqrt{c}=49-45=4\)

<=> c = 16

=> \(c^2-21=16^2-21=16.\left(16-1\right)-5=15.16-5=240-5=235\)

Call Sara's marbles is a

        Jo's marbles is b

We have :

\(\left\{{}\begin{matrix}a+3=2\left(b-3\right)\\b+2=2\left(a-2\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+3=2b-6\\b+2=2a-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2b-9\\b=2a-6\end{matrix}\right.\)

Change a = 2b - 9 into b = 2a - 6

=> a = 2.(2a - 6) - 9

=> a = 4a - 12 - 9 

=> 21 = 3a

=> a = 7

=> b = 8

So the number of Sara's marbles are 7

and  Jo's marbles are 8 

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

+) Apply inequality \(\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a,b,m\in N^+\right)\)

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

From (1) and (2)

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

Denote  \(C=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}\)

+) We have :

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

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

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

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

So \(C>\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}=\dfrac{a+b+c+d}{a+b+c+d}=1\)

+) We have this too :v 

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

\(\dfrac{b}{b+c+d}< \dfrac{b}{b+d}\)

\(\dfrac{c}{c+d+a}< \dfrac{c}{c+a}\)

\(\dfrac{d}{d+a+b}< \dfrac{d}{b+d}\)

Plus equality to equality :

\(\Rightarrow C< \dfrac{a}{a+c}+\dfrac{b}{b+d}+\dfrac{c}{a+c}+\dfrac{d}{b+d}=\dfrac{a+c}{a+c}+\dfrac{b+d}{b+d}=2\) 

Finally , \(1< C< 2\) :) 

If you want to do shorter than that way , we can do this:

\(2^{3\left(x+1\right)}=4^{x+16}\)

\(\Leftrightarrow2^{3\left(x+1\right)}=2^{2\left(x+16\right)}\)

\(\Leftrightarrow3\left(x+1\right)=2\left(x+16\right)\)

\(\Leftrightarrow3x+3=2x+32\)

\(\Leftrightarrow x=29\)

This way can be used in questions , but in a few question, you shouldn't do this , because it can't find all the results. So that, you will wrong  

I recommend using that way.

2^{3(x + 1)} = 4^{x + 16}

2^{3(x + 1)} = 2^{2(x + 16)}

2^{3x + 3} - 2^{2x + 32} = 0

2^{2x + 3} . 2^x - x^{2x + 3} . 2²⁹ = 0

2^{2x + 3} . (2^x - 2²⁹) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2^{2x+3}=0\\2^x-2^{29}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}error\\2^x=2^{29}\end{matrix}\right.\Rightarrow x=29\)

I'm sorry , i'm wrong at question c !

We have : 

2⁵ = 32 = 2^x

=> x = 5