Because : BD // AE

=> \(\widehat{DBA}=\widehat{BAE}\)  (2 Alternate interior angles)

In another face , we got : \(\widehat{DBA}=\widehat{BEA}\) (2 Corresponding angles)

So \(\widehat{BAE}=\widehat{BEA}\)

Following Pythagoras theorem , the length of the diagonal is :

\(Diagonal=\sqrt{4^2+4^2}=\sqrt{32}=6\) (cm)

On the 6th line , change to :

\(\Delta ABC\) and \(\Delta ADE\) are similar triangles. 

We have :

\(BC\perp AD\)

\(DE\perp AD\)

=> BC // DE

But BC cut AD at B and cut AE at C

So \(\Delta ABC~\Delta ADE\)

=> \(\dfrac{AB}{AD}=\dfrac{AC}{AE}=\dfrac{BC}{DE}\)

\(\Leftrightarrow\dfrac{4}{AD}=\dfrac{5}{AE}=\dfrac{BC}{9}\)

Following Pythagoras theorem , we got :

\(BC=\sqrt{AC^2-AB^2}=\sqrt{25-16}=3\)

So \(\dfrac{4}{AD}=\dfrac{5}{AE}=\dfrac{3}{9}=\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}AD=12\\AE=15\\DC=9\end{matrix}\right.\)

\(\Rightarrow P_{\Delta ADE}=12+15+9=36\left(cm\right)\)

\(\dfrac{1}{x}+\dfrac{1}{2x}=\dfrac{4}{5}\)

\(\dfrac{2}{2x}+\dfrac{1}{2x}=\dfrac{4}{5}\)

\(\dfrac{3}{2x}=\dfrac{4}{5}\)

\(\Leftrightarrow3.5=4.2x\)

\(\Leftrightarrow15=8x\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

@AL No.....It's true ! :v 

e) x³ - 2x²y + 3xy² 

= x³ - 2x²y + 6xy² - 3xy²

= x³ - 3xy² - 2x²y + 6xy²

= x(x² - 3y²) - 2y(x² - 3y²)

= (x - 2y)(x² - 3y²)

f) x² − xy − 2x + 2y

= x(x - y) - 2(x - y)

= (x - 2)(x - y) 

"Phân tích đa thức thành nhân tử" ??? Congratulation , you are super real ! :v 

a) x³ - x² + x - 1

= x²(x - 1) + (x - 1)

= (x² + 1)(x - 1)

b) 6x²y - 2xy² + 3x - y

= 2xy(3x - y) + (3x - y)

= (2xy + 1)(3x - y)

c) 4x² + 1 (It's wrong)

Change to : 4x² - 1

= (2x)² - 1

= (2x + 1)(2x - 1)

d) x² - 9x + 8

= x² - x - 8x + 8

= x(x - 1) - 8(x - 1)

= (x - 8)(x - 1)

Ohhh.....i'm sorry about that @AL :)

Change to : \(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\)

Call intersection of C and BH is W

Revised to : Call intersection of EC and BH is W

Are you kidding me ??? :V

a) Have : \(\widehat{EAC}=90^0+\widehat{BAC}\)

               \(\widehat{BAH}=90^0+\widehat{BAC}\)

=> \(\widehat{EAC}=\widehat{BAH}\)

Consider \(\Delta EAC\) and \(\Delta BAH\) , have :

\(\widehat{EAC}=\widehat{BAH}\)

EA = EB (From hypothesis)         => \(\Delta EAC\) = \(\Delta BAH\)

HA = HC (From hypothesis) 

=> EC = BH

Call intersection of C and BH is W

Similar, call intersection of BH and AC is O

We have :

            \(\widehat{HAO}+\widehat{AOH}+\widehat{OHA}=180^0\) (total three angle in a triangle)

And :   \(\widehat{CWO}+\widehat{WOC}+\widehat{OCW}=180^0\)

We got : \(\widehat{AOH}=\widehat{WOC}\) 

              \(\widehat{OHA}=\widehat{OCW}\) (From 2 equal triangles)

=> \(\widehat{HAO}=\widehat{CWO}=90^0\) 

=> \(HB\perp EC\)

b) We have :

NI is the average line of \(\Delta BHC\)

=> \(NI=\dfrac{1}{2}BH\)     and     NI // BH

Similar with MI we have :

\(MI=\dfrac{1}{2}EC\)          and       MI // EC

But BH = EC

=> MI = NI              (1)

We have :

\(HB\perp EC\)

EC // MI 

=> \(BH\perp MI\)

Similar : 

\(BH\perp MI\)

NI // BH

=> \(MI\perp NI\)          (2)

From (1) and (2)

=> \(\Delta MIN\) is the isosceles square triangle at I 

Following threads , we got :

EP // AF (Ex // AF)

FP // AE (Fy // AE)

=> AEPT is the parallelogram  (1)

<=> AF = AE

With ABCD is the square , we have : AB = AD

Consider \(\Delta ADF\) and \(\Delta ABE\) , have :

AB = AD

AE = AF                                              => \(\Delta ADF\) = \(\Delta ABE\) (e - e - e)

BE = DF (From the hypothesis) 

=> \(\widehat{FAD}=\widehat{EAB}\)

We have :

\(\widehat{EAB}+\widehat{DAE}=90^0\)

but \(\widehat{FAD}=\widehat{EAB}\)

=> \(\widehat{FAD}+\widehat{DAE}=90^0\)        (2)

From (1) and (2)

=> AEPF is the square 

Let B = \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+......+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+........+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+.....+\dfrac{1}{99}+\dfrac{1}{100}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+........+\dfrac{1}{100}\)  

So \(\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+.......+\dfrac{1}{100}\right):\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}\right)=1\)

\(\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}......\dfrac{2499}{2500}\)

\(=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}......\dfrac{49.51}{50.50}\)

\(=\dfrac{2.4.3.5.4.6....49.51}{3.3.4.4.5.5.....50.50}\)

\(=\dfrac{2.4.5.6....51}{3.4.5....50}\)

\(=\dfrac{2.51}{3.50}\)

\(=\dfrac{17}{25}\)

We have :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{100!}=\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+....+\dfrac{1}{1.2.....100}\)

\(.........< \dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{99.100}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(.........< 1-\dfrac{1}{100}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}< 1\)

We got : 

x = 9

=> x + 1 = 10

We have :

A = x²⁰¹⁶ - 10.x²⁰¹⁵ + 10.x²⁰¹⁴ - 10.x²⁰¹³ + ........ + 10x² - 10x + 2026

A = x²⁰¹⁶ - (x + 1).x²⁰¹⁵ + (x + 1).x²⁰¹⁴ - (x + 1).x²⁰¹³ + ........ + (x + 1).x² - (x + 1)x + 2026

A = x²⁰¹⁶ - x²⁰¹⁶ - x²⁰¹⁵ + x²⁰¹⁵ + x²⁰¹⁴ - x²⁰¹⁴ - x²⁰¹³ + ......... + x³ + x² - x² - x + 2026

A = -x + 2016

A = -9 + 2026

A = 2017

We got : 

x = 9

=> x + 1 = 10

We have :

A = x²⁰¹⁶ - 10.x²⁰¹⁵ + 10.x²⁰¹⁴ - 10.x²⁰¹³ + ........ + 10x² - 10x + 2026

A = x²⁰¹⁶ - (x + 1).x²⁰¹⁵ + (x + 1).x²⁰¹⁴ - (x + 1).x²⁰¹³ + ........ + (x + 1).x² - (x + 1)x + 2026

A = x²⁰¹⁶ - x²⁰¹⁶ - x²⁰¹⁵ + x²⁰¹⁵ + x²⁰¹⁴ - x²⁰¹⁴ - x²⁰¹³ + ......... + x³ + x² - x² - x + 2026

A = -x + 2016

A = -9 + 2026

A = 2017

Call F₁ is an exterior angle 

 and F₂ is an adjacent angle 

We have :

F₁ + F₂ = 180⁰ ( 2 adjacent offset angle )

=> F₂ = 180⁰ - 5x - 34⁰ = 146⁰ - 5x

Consider triangle DEF , we have :

F₂ + D + E = 180⁰ - 5x + 2x + 5⁰ + 4x + 20⁰ = 180⁰

<=> 205⁰ + x = 180⁰

=> x = 205⁰ - 180⁰

=> x = 25⁰

Edit post : Give 2017 positive integer numbers a1 , a2 , a3 , a4 , ............... a2017 so that 

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Suppose a₁ , a₂ , a₃ , a₄ , ...... , a₂₀₁₇ are different .

And 0 < a₁ < a₂ < .............. < a₂₀₁₇

Since is a positive integer so that we have :

\(\dfrac{1}{a_1}+\dfrac{1}{a_2}+........+\dfrac{1}{a_{2017}}\le\dfrac{1}{1}+\dfrac{1}{2}+.........+\dfrac{1}{2017}< \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{2}...............+\dfrac{1}{2}=1+\dfrac{2016}{2}=1+1008=1009\)

From this , if those numbers are different so that the sum always smaller than 1009 , but those are equal to 1009 , that means there are 2 in 2017 numbers are equal.

ohhh.....! I'm sorry , the last row will be :

So , there are 120 kilometres do you live from the station .