Given: a + b + 2c = 1.

Find the minimum and maximum of A = ab - 4bc - 4ca

a - a always equal 0

The correct answer is C

5. Given the triangle ABC balance at A (AB = AC). On BC take M any such that BM < CM. From M draw parallel lines with AC cut AB at E and parallel with AB cut AC at F. Let N be the symmetry point of M over EF.
a) Calculate the perimeter of AEMF. Know: AB = 7cm
b) Prove that: AFEN is a trapezoid trait
c) Calculated: \(\widehat{ANB}+\widehat{ACB}\)
d) Where M is the AEMF quadrilateral is diamond and needs to be added to the triangle ABC so that the AEMF is square.

4. (easy) Via the center of the triangle G, the straight line parallel to AC, cut AB and BC respectively at M and N. Calculate the length of the field, know AM + NC = 16 (cm); Triangle triangle ABC equals 75 cm

3. Given triangle ABC. On BC, CA, AB take D, E, F so that AD, BE, DF intersect at H.

Prove that:

a, \(\dfrac{AH}{AD}+\dfrac{BH}{BE}+\dfrac{CH}{CF}=2\)

b, \(\dfrac{AH}{HD}+\dfrac{BH}{HE}+\dfrac{CH}{HF}\ge6\)

2. Given triangle ABC. M is a point in the interior of triangle ABC. Let D, E, F be the midpoints AB, AC, BC; A ', B', C 'is the symmetry point of M through F, E, D.
a, CMR: AB'A'B is a parallelogram.
b, CMR: CC 'goes through the midpoint of AA'

In this web - Mathulike mostly is algebraic and only few geometry. So I will post some exercises of geometry.

1. Given ABCD rectangle with AB = 2AD, call E, I in turn is the midpoint of AB and CD. Connect D to E. Draw the Dx beam perpendicular to DE, the Dx beam to the opposite ray of the CB ray at M. On the opposite side of the CE beam take the K point so that DM = EK. Let G be the intersection of DK and EM.
a) Calculate DBK angle.
b) Let F be the right-angled foot from K to BM. Prove that four points A, I, G, H are in the same line.

We have: 

\(a^2+b^2+c^2\ge ab+bc+ca\)

The "=" happen when and only when a = b = c

Applying the Cauchy-Schwarz inequality, we have:

\(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\le\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2+2+2}\le\dfrac{9}{ab+bc+ca+6}=\dfrac{9}{9}=1\)

So......................

\(\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{1}{\dfrac{17}{4}}}}=\dfrac{1}{4+\dfrac{1}{4+\dfrac{4}{17}}}=\dfrac{1}{4+\dfrac{1}{\dfrac{72}{17}}}=\dfrac{1}{4+\dfrac{17}{72}}=\dfrac{1}{\dfrac{305}{72}}=\dfrac{72}{305}\)

\(\left(\left(X^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}\right)^{\dfrac{1}{2}}=\sqrt{\sqrt{\sqrt{X}}}=\sqrt[2^3]{X}=\sqrt[8]{X}\)

The fraction in odd location have the denominator is sequence of odd numbers from 3

The fraction in even location have the denominator is sequence of 2ⁿ from 2¹

And their numerator is 1

So the fraction in ? is \(\dfrac{1}{13}\)

Their average is: \(\dfrac{6+7+42}{16}=\dfrac{55}{16}=3,4375\)

\(\dfrac{30\%\cdot\dfrac{25}{17}}{\dfrac{3}{4}\cdot\dfrac{1}{2}}=\dfrac{\dfrac{15}{34}}{\dfrac{3}{8}}=\dfrac{20}{17}\)

200 days = 28 weeks and 4 days

So 200 days from now will be 3 + 4 = 7 or Saturday 

Call a zeeko is x, a teeko is y. We have:

\(\dfrac{1}{2}x=60\%y\)

\(\Leftrightarrow0,5x=0,6y\)

\(\Leftrightarrow x=1,2y\)

\(\Leftrightarrow\dfrac{x}{y}=1,2\)

With this post, we have:

130%A = B

60%B = C

120%C = D

=> D = 60% of 120% of B

=> D = 72%B

\(\dfrac{1}{10\sqrt{10}}=\dfrac{1}{\sqrt{10^3}}\)

\(\dfrac{\sqrt{10}}{\sqrt{10^4}}=\dfrac{1}{\sqrt{10^3}}\)

So A = B

The answer is c

\((x-1)(2x-1) = 9-x\)

=> \(2x^2 - 3x + 1 = 9- x\)

=> \(2x^2 - 2x = 8\)

=> \(2x^2 - 2x - 8 = 0\)

=> \(2(x^2 - x - 4)= 0\)

=> \(2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{17}{2}=0\)

=> \(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{2}\)

=> \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)

=> \(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

3rd term is: 1 + 2 = 3

4th term is: 1 + 2 + 3 = 6

5th term is: 1 + 2 + 3 + 6 = 12

....

15th term is: 1 + 2 + 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 = 1536 

\(0.2\overline{36}=\dfrac{236-2}{990}=\dfrac{234}{990}=\dfrac{13}{55}\)

In recent times, I have seen some questions not related to mathematics. And that is not good. I suspect there are some accounts where the main account is https://e-learning.codienhanoi.edu.vn/profile/ducanh2007 has posted the question itself and tick yourself. As an coodinator, I hope Admin will soon resolve this issue. And I advise you to have the nickname "Cristiano Ronaldo" stop ticking yourself. I hope Mr.Bee will solve it satisfactorily.

And to stop this status, I will post some question in my free time. 

People call this triangle is Pascal.

                                                  1

                                        1                    1

                              1                    2                    1

                    1                    3                    3                    1

          1                    4                    6                    4                    1

1                    5                    10                    10                    5                    1

So the answer is 1,5,10,10,5,1

This sequence is ...,5k,6k,7k,8k,9k,10k,11k,12k,... with k = 4

So each terms in this will \(⋮4\)

So in 4 answers, answer d is correct :]

This sequence is 0k,1k,2k,3k,4k,5k,6k,7k,8k,...,300k with k = 3

So the missing numbers in "..." is 9k to 299k

So there are: 299 - 9 + 1 = 291 numbers in this

\(S_{ABC}=\dfrac{\left(13-11\right)\left(17-2\right)}{2}=\dfrac{2\cdot15}{2}=15\left(unit^2\right)\)

Way 1: \(\left\{{}\begin{matrix}GCD\left(48;72\right)=24\\LCM\left(48;72\right)=144\end{matrix}\right.\)

So their product is: 24 . 144 = 3456

Way 2: 

Their product is: 48 . 72 = 3456

\(f\left(x\right)=3\Rightarrow\sqrt{x+4}=3\Rightarrow x+4=9\Rightarrow x=5\)

If A true 

=> B false and C true 

=> D false and A false (Contrary to suppose) 

So A false 

=> C is true statement.

a # b = a + 2b 

=> 3 # 4 = 3 + 2.4 = 3 + 8 = 11

It takes \(\dfrac{35}{30}=\dfrac{7}{6}\left(h\right)=70'\) to drive all this distance at 30mi/h

It takes \(\dfrac{25}{30}=\dfrac{5}{6}\left(h\right)=50'\)to drive all this distance at 25mi/h

So drive at 30mi/h is takes fewer than 25mi/h is: 70' - 50' = 20'

Answer: 20 minutes

\(2015=5\cdot13\cdot31\)

So their sum is: 5 + 13 + 31 = 49

Tell the number to find is x

We have:

x : 8 remainder 7

x : 9 remainder 8

x : 12 remainder 11

==> x + 1 ⋮ 7,8,11

\(\Rightarrow x+1\in CM\left(7,8,11\right)\)

But x is the smallest number 

=> \(x+1\) is \(LCM\left(7,8,11\right)=616\) (Get out 0 because x is positive number) 

So x = 615  

From the 25th day to 284th day has: \(284-25+1=260\left(days\right)\)

   260 days = 37 weeks and 1 day.

So the 284th day is Sunday

The value of (1 + 2 + 3) + (2 + 3 + 4) + (3 + 4 + 5) + … + (38 + 39 + 40) + (39 + 40 + 41) is:

1 + 2.(2 + 3 + ... + 40) + 41 = 42 + 1638 = 1680

\(9x^2+4y^2=20xy\)

\(\Leftrightarrow\left(3x\right)^2-12xy+\left(2y\right)^2=8xy\)

\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)

\(9x^2+4y^2=20xy\)

\(\Leftrightarrow\left(3x\right)^2+12xy+\left(2y\right)^2=32xy\)

\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)

\(\Rightarrow p=\sqrt{\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}}=\sqrt{\dfrac{8xy}{32xy}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-2\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(x+y+z\ne0\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)

\(\Rightarrow p=\dfrac{x^{2018}+1}{x^{2018}+x^{2018}+x^{2018}+3}=\dfrac{x^{2018}+1}{3\left(x^{2018}+1\right)}=\dfrac{1}{3}\)