\(0,6\left(x+10\right)+0,4\left(11x-5\right)=0,7x+25,5\)

\(0,6.x+6+0,4.11x-2=0,7x+25,5\)

\(\left(0,6x+4,4x\right)-0,7x=25,5-4\)

\(\left(0,6+4,4-0,7\right)x=21,5\)

\(4,3x=21,5\)

\(\Rightarrow x=5\)

\(A=\dfrac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\dfrac{1}{5}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)

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

\(=\dfrac{1}{1}-\dfrac{1}{6}=\dfrac{5}{6}\)

10.2 + 10.3 = 20.5

4.3 + 2.4 = 6.7

Denote : \(A=a_1^2+a_2^2+....+a_n^2\)

              \(B=b_1^2+b_2^2+.....+b_n^2\)

              \(C=\left(a_1b_1+a_2b_2+........+a_nb_n\right)^2\)

Need to prove : AB \(\ge\) C²

If A = 0 so that \(a_1=a_2=....=a_n\), inequality is proven. Similar with B = 0 , so we need to consider if A and B different to 0

With every x , we have got :

\(\left(a_1x-b_1\right)^2\ge0\Rightarrow a^2_1.x^2+2.a_1b_1x+b_1^2\ge0\)

\(\left(a_2x-b_2\right)^2\ge0\Rightarrow a^2_2.x^2+2.a_2b_2x+b_2^2\ge0\)

.............................................

\(\left(a_nx-b_n\right)^2\ge0\Rightarrow a^2_n.x^2+2.a_nb_nx+b_n^2\ge0\)

=> \(\left(a_1^2+a_2^2+........+a_n^2\right).x^2-2.\left(a_1b_1+a_2b_2+.....+a_nb_n\right)x+\left(b_1^2+b_2^2+.......+b_n^2\right)\ge0\)(1)

That mean \(A.x^2-2.C.x+B\ge0\)

Because (1) right with all x so that , change x = \(\dfrac{C}{A}\) into (1) , we have :

\(A.\dfrac{C^2}{A^2}-2.\dfrac{C^2}{A^2}+B\ge0\)

\(\Rightarrow AB-C^2\ge0\)

\(\Rightarrow AB\ge C^2\)

=> \(\left(a_1^2+a_2^2+....+a_n^2\right)\). \(\left(b_1^2+b_2^2+..........+b_n^2\right)\) \(\ge\left(a_1b_1+a_2b_2+......+a_nb_n\right)^2\)

@Dao Trong Luan , you are wrong :v 

\(A=1+2+2^2+..........+2^{14}\)

\(A=\left(2^0+2^1+2^2+2^3+2^4\right)+\left(2^5+2^6+......+2^9\right)+\left(2^{10}+2^{11}+......+2^{14}\right)\)

\(A=31.1+2^5.31+2^{10}.31\)

\(A=31\left(1+2^5+2^{10}\right)⋮31\)

Connect A and C 

Following total 3 corners in a triangle, we have :

\(\widehat{ADC}+\widehat{DCA}+\widehat{CAD}=180^0\)  (\(\Delta ACD\))

\(\widehat{ACB}+\widehat{CBA}+\widehat{BAC}=180^0\)  (\(\Delta ABC\))

=> \(\widehat{ADC}+\widehat{DCA}+\widehat{CAD}\) + \(\widehat{ACB}+\widehat{CBA}+\widehat{BAC}\) = 180⁰ + 180⁰ = 360⁰

<=> \(\widehat{ADC}+\widehat{DCB}+\widehat{CBA}+\widehat{BAD}=360^0\)

At a speed of 50km/h, the time to complete is : 

\(t=\dfrac{s}{t}=\dfrac{900}{50}=18\left(hour\right)\)

Change : 900 km = 900. 0,62137 = 559.233 miles 

So , the velocity calculated by miles/hour is : \(\dfrac{599.233}{18}=33.29.....\approx33.291\)  (miles/hour)

With the time complete in 3 hours, the velocity is : \(\dfrac{599.233}{3}=199.744\) (miles/hour)

Hence, there are 199.744 - 33.291 = 166.453 (miles/hour) faster you must average while driving.

Apply Pythagorean theorem , we have :

\(\text{Hypotenuse}=\sqrt{6^2+7^2}=\sqrt{85}\left(cm\right)\)

1.2.3....100 = 100! = ..........0

So , the last digit of this product is 0

\(\dfrac{x}{2}+x+\dfrac{x}{3}+\dfrac{x}{4}=x.\left(\dfrac{1}{2}+1+\dfrac{1}{3}+\dfrac{1}{4}\right)=25\)

\(\dfrac{x.25}{12}=25\)

=> x = 12 

a) We have : DM = HD

                     BD = CD

               \(\widehat{BDM}=\widehat{CDH}\)  ( 2 Vertical angles )

=> \(\Delta BDM=\Delta CDH\)   (1) 

b) From (1) , we have :

\(\widehat{HCB}=\widehat{MBC}\)

But \(\Delta ABC\) is the isosceles triangle

So that \(\widehat{HCB}=\widehat{ABC}\)

=> \(\widehat{MBC}=\widehat{ABC}\)

=> BC is the bisector of \(\widehat{ABM}\)

c) I don't understand the thread 

Use Zigma to calculator the sum A , we have :

\(\sum\limits^{105}_{x=0}\left(1+X\right)=5670\)

So A = 5670 

Perhaps Bunyakovsky's inequality :v , Sure right =)))

We have :

\(n.\left(a_1^2+a_2^2+.....+a_n^2\right)=\left(1+1+1+.......+1\right).\left(a_1^2+a_2^2.......+a_n^2\right)\)

                                                ........ n number 1..........

\(L.H.S\ge\left(1.a_1+1.a_2+.....+1.a_n\right)^2=\left(a_1+a_2+.....+a_n\right)^2\)

Done , ok !

a) \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(a^2+2ab+b^2\le2a^2+2b^2\)

\(a^2-2ab+b^2\ge0\)

\(\left(a-b\right)^2\ge0\)

b) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(a^2+b^2+c^2+2ab+2bc+2ca\le3a^2+3b^2+3c^2\)

\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

:( another way 

We going to prove this inequality : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

<=> \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

<=> \(\left(a+b\right)^2\ge4ab\)

<=> \(a^2+2ab+b^2\ge4ab\)

<=> \(a^2-2ab+b^2\ge0\) 

<=> \(\left(a-b\right)^2\ge0\) (it's true)

So that : 

\(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\le\dfrac{1}{\dfrac{4}{a+b}}+\dfrac{1}{\dfrac{4}{b+c}}+\dfrac{1}{\dfrac{4}{c+a}}\)

\(..............\le\dfrac{a+b}{4}+\dfrac{b+c}{4}+\dfrac{c+a}{4}=\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)

When a = b = c 

@Phan Thanh Tinh , what about this way :

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)

\(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)

\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{x}.\dfrac{1}{y}.3\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{-1}{z^3}\)

\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-\dfrac{1}{x}.\dfrac{1}{y}.3\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-\dfrac{1}{x}.\dfrac{1}{y}.3.\dfrac{-1}{z}=-\dfrac{3}{xyz}\)

Have : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow yz+zx+xy=0\)

We have : 

\(AB\perp DM\) at O

OD = OM

=> \(\Delta ADM\) is a isosceles triangle at A (1)

Similar with \(\Delta AME\)

=>  \(\Delta ADM\) is a isosceles triangle at A (2)

From (1) and (2) 

=> AD = AE = AM
 

No, it's my mistake , not you :V 

There is something is wrong here 

Change to :

We have \(a^2+b^2\ge5\)

=> \(9-\left(a^2+b^2\right)\le4\)

=> \(2ab\le4\)

=> \(ab\le2\)

<=> \(a^2b^2\ge4\)

=> \(4a^2b^2\ge16\)