Lightning Farron:I don't understand

Lightning Farron:\(2\left(4x+1\right)+\dfrac{108x-73}{\left(5-12x\right)\sqrt{3x-2}-20x+13}>0\).Why?

Lê Quốc Trần Anh that's right.It used many users to swear for me.

I used to remind but it unheard.

KEITA FC 8C:MAY KHONG NEN TU HOI TU TRA LOI

MAY LAM THE KHONG DUOC GI DAU

A hour Pump Q can fill a water is:

           \(1:15=\dfrac{1}{15}\)(tank)

A hour Pump P can fill a water is:

        \(1:12=\dfrac{1}{12}\)(tank)

A hour Pump P and Pump Q can fill a water is:

    \(\dfrac{1}{12}+\dfrac{1}{15}=\dfrac{3}{20}\)(tank)

The time for Pump P and Pump Q fill 60% full is:

   \(\dfrac{3}{5}:\dfrac{3}{20}=4\)(hours)

The time for Pump Q fill 40% full is:

         \(\dfrac{2}{5}:\dfrac{1}{15}=6\)(hours)

The number of hour for it take to completely fill the tank is:

   6+4=10(hours)

Answer:10 hours

With D=1 \(\Rightarrow\)21\(\times\)151=8071\(\Rightarrow3171=8071\)(unsatisfactory)

With D=3\(\Rightarrow23\times351=8073\Rightarrow8073=8073\)(satisfy)

With D\(\ge5\)\(\Rightarrow2D\times D51\ge25\times551=13775>8073\)(unsatisfactory)

Apply extend Bunyakovsky inequality ,we have:

\(a^3+b^3+c^3\)\(=\dfrac{a^4}{a}+\dfrac{b^4}{b}+\dfrac{c^4}{c}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a+b+c}=\dfrac{81}{a+b+c}\)                                              (1)                                      

Other way:

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\Rightarrow27\ge\left(a+b+c\right)^2\)\(\Rightarrow a+b+c\le\sqrt{27}=3\sqrt{3}\)                 (2)

From (1) and (2)\(\Rightarrow a^3+b^3+c^3\ge\dfrac{81}{a+b+c}\ge\dfrac{81}{3\sqrt{3}}=9\sqrt{3}\)

\(\Rightarrow min_M=9\sqrt{3}\) when a=b=c=\(\sqrt{3}\) because a,b,c\(\ge-1\)

 KEITA FC 8C,Fc Alan Walker:Tao khuyen may khong nen tu hoi tu tra loi nhu the

We have:\(\left\{{}\begin{matrix}a+b=11\\a.b=24\end{matrix}\right.\)\(\Rightarrow ab+a+b=35\Rightarrow ab+a+b+1=36\Rightarrow\left(a+1\right)\left(b+1\right)=36\)

We have table:(because a,b same kind so not consider a,b negative)

So the absolute difference between those two number is:\(\left|8-3\right|=\left|5\right|=5\)

Put the number of nikels,dimes and quarters is:a,b,c

We have:\(\left\{{}\begin{matrix}a+b+c=37\\a-b=4\\c-a=2\end{matrix}\right.\)\(\Rightarrow\left(a+b+c\right)+\left(a-b\right)+\left(c-a\right)=43\)\(\Rightarrow a+2c=43\).

Because \(c-a=2\) so \(\left(a+2c\right)+\left(c-a\right)=45\Rightarrow3c=45\Rightarrow c=15\Rightarrow\left\{{}\begin{matrix}a=13\\b=9\end{matrix}\right.\)

Answer:The number of nickels is 13

              The number of dimes is:9

              The number of quarters is:15

This is a rectangular but you draw wrong.

You must draw some line equal cut line.

You draw shorter line than right picture.

*,There are 4 Sunday days in April

*,Tomorow is:30th                         Yesterday is:28th

*,The Sunday week yesterday is:12-7=5th

Because x+y creates 1 part 2 of a diagonal line of the square so \(x+y=\sqrt{\dfrac{4}{2}}=\sqrt{2}\)

So x+y=\(\sqrt{z}\Rightarrow\sqrt{2}=\sqrt{z}\Rightarrow z=2\)

Answer:The value of z is:2

Chúng ta có: x + y =\(\sqrt{\dfrac{4}{2}}=\sqrt{2}\) because x+y creates 1 part 2 of a diagonal line

So x+y=\(\sqrt{z}\Rightarrow\sqrt{2}=\sqrt{z}\)\(\Rightarrow z=2\)

Answer:z=2

The first number:3=1.3

The second number:6=3.2

The third number:9=3.3

.........

So the 99th number of row is:

   3.99=297

Answer:297

The side of the hexagon is:

   \(1\times2\div2=1\)(unit)

The perimeter of a semicircle is:

   \(\dfrac{1\times3,14}{2}=1,57\)

The perimeter of the figure formed by these semicircle is:

      \(1,57\times6=9,42\approx9,4\)

Answer:9,4

We have:\(99.45.15.25+13-\left(25.99.3.5.9.5\right)\)

\(=99.45.15.25+13-25.99.15.45\)

\(=13\)

Answer:13

Put the edge of square is:x

\(\pi\) have a unit is inch

 The area of the big circle is:

   12 x 12 =144(\(\pi\))

The edge of square (or diameter of the small circle) is:

\(x=\sqrt{\dfrac{\left(12.2\right)^2}{2}}=\sqrt{\dfrac{576}{2}}=\sqrt{288}\)(inches)

The area of the small circle is:

   \(\left(\dfrac{\sqrt{288}}{2}\right)^2=72\left(\pi\right)\)

The area of the shaded region between the two circles is:

   144-72=72\(\left(\pi\right)\)

Answer:72\(\pi\)

The length of diamter AB is

  \(AB=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}\)

The area of a circle that has diameter is:

   \(\left(\dfrac{\sqrt{52}}{2}\right)^2.3,14\)\(=\dfrac{52}{4}\pi=13\pi\)

Answer:13\(\pi\)

We have:\(n^3-1\equiv-1\)(mod n)

\(\Rightarrow\left(n^3-1\right)^{111}\equiv\left(-1\right)^{111}\)(mod n)

\(\Rightarrow\left(n^3-1\right)^{111}\equiv-1\)(mod n)                                                                   (1)

Other way:\(n^2-1\equiv-1\)(mod n)

\(\Rightarrow\left(n^2-1\right)^{333}\equiv\left(-1\right)^{333}\)(mod n)

\(\Rightarrow\left(n^2-1\right)^{333}\equiv-1\)(mod n)                                                                   (2)

From (1) and (2)\(\Rightarrow\left(n^3-1\right)^{111}.\left(n^2-1\right)^{333}\equiv\left(-1\right).\left(-1\right)\)(mod n)

\(\Rightarrow\left(n^3-1\right)^{111}.\left(n^2-1\right)^{333}\equiv1\)(mod n)

So the surplus when divide (n³-1)¹¹¹.(n²-1)³³³ for n is:1