1 + 3 + .... + 1000 = (1 + 3 + 5 + ..... + 999) + 1000

Apply the formula to calculator the total number , we have :

\(\dfrac{\left[\left(999-1\right):2+1\right].\left(999+1\right)}{2}=\dfrac{550.1000}{2}=550.500=275000\)

=> 275000 + 1000 = 276000

(x + 2).(x + 2) = 24

(x + 2)² = 24

\(\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{24}\\x+2=-\sqrt{24}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{24}-2\\x=-\sqrt{24}-2\end{matrix}\right.\)

So x = \(\sqrt{24}-2\) or x = \(-\sqrt{24}-2\)

a) With 2x + 3 = x + 2

=> 2x - x = 2 - 3

=> x = -1

With 2x + 3 = -(x + 2)

=> 2x + 3 = -x - 2

=> 2x + x = -2 - 3

=> 3x = -5

=> x = \(\dfrac{-5}{3}\)

So x = -1 and x = \(\dfrac{-5}{3}\)

b) We have : 

A = |x - 2006| + |2007 - x| \(\ge\) |x - 2006 + 2007 - x| = |1| = 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2006\right|\le0\\\left|2007-x\right|\le0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2006\\x=2007\end{matrix}\right.\)

So when x = 2006 ; 2007 then value of A smallest 

We have :

x < -0.8 \(\Rightarrow\) x + 0,8 < 0 \(\Rightarrow\)|x + 0.8| = -(x + 0.8)

x < -0.8 \(\Rightarrow\) x < 25 \(\Rightarrow\) |x - 25| = 25 - x

\(\Rightarrow\) A = -(x + 0.8) - (25 - x) + 1.9

     A = -x - 0.8 - 25 + x + 1.9

     A = -23.9 

We have : 

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

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

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

+> If a + b + c + d \(\ne0\)

=> a = b = c = d (Because the same of namerator)

=> M = 1 + 1 + 1 + 1 = 4

+> If a + b + c + d = 0

=> a + b = -(c + d)

      b + c = -(a + d)

      c + d = -(a + b)

      a + d = -(b + c)

Change this into M , we have :

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

\(M=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(a+d\right)}{a+d}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

So M = 4 and M = -4

We have : 

\(\dfrac{bt-cn}{a}=\dfrac{cm-at}{b}=\dfrac{an-bm}{c}\)

\(\Rightarrow\dfrac{a\left(bt-cn\right)}{a^2}=\dfrac{b\left(cm-at\right)}{b^2}=\dfrac{c\left(an-bm\right)}{c^2}\)

\(\Rightarrow\dfrac{abt-acn}{a^2}=\dfrac{bcm-bat}{b^2}=\dfrac{acn-bcm}{c^2}=\dfrac{abt-acn+bcm-bat+acn-bcm}{a^2+b^2+c^2}=0\)

\(\Rightarrow\left\{{}\begin{matrix}bt-cn=0\\cm-at=0\\an-bm=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}bt=cn\Rightarrow\dfrac{n}{b}=\dfrac{t}{c}\\cm=at\Rightarrow\dfrac{t}{c}=\dfrac{m}{a}\\an=bm\Rightarrow\dfrac{m}{a}=\dfrac{n}{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{m}{a}=\dfrac{n}{b}=\dfrac{t}{c}\)

Give : \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)

Change x,y,z into 5z - 3x - 4y = 50 , we have :

5(6k + 5) - 3(2k + 1) - 4(4k - 3) = 50

30k + 25 - 6k - 3 - 16k + 12 = 50 

8k + 34 = 50

8k = 16

=> k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=2k+1=2.2+1=5\\y=4k-3=4.2-3=5\\z=6k+5=6.2+5=17\end{matrix}\right.\)

My drawing is not accurate ha ! ^ ^ Sorry about that !

Draw BX parallel to AC

=> \(\widehat{BXE}=\widehat{CEX}\)

Consider \(\Delta BMX\) and \(\Delta CEM\) , we have :

BM = CM

\(\widehat{BMX}=\widehat{CME}\)  => \(\Delta BMX\) = \(\Delta CEM\)

\(\widehat{BXE}=\widehat{CEX}\)

Consider \(\Delta AHD\) and \(\Delta AHE\) , we have :

\(\widehat{DAH}=\widehat{CAH}\)

AH general                     => \(\Delta AHD\) = \(\Delta AHE\)

\(\widehat{AHD}=\widehat{AHE}=90^0\)

=> \(\widehat{ADE}=\widehat{AED}\)

But \(\widehat{AED}=\widehat{BXD}\)

=> \(\widehat{ADE}=\widehat{BXD}\)

=>\(\Delta\)BDX  isosceles 

=> BD = BX

But BX = CE

=> BD = CE

you lost the very important rules , i don't want to talk to you anymore .

If you say i am wrong about 2017 , i agree , but about you , from where do you have \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\) 

About what , show me !Love the Name of Jiang

Love the Name of Jiang . Stop rewrite my anwser , you will be bad if you still do that !

Applying the same sequence properties, we have

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

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)

Change a = b = c , i have :

a + b + c = a + a + a = 2017 

=> 3a = 2017

=> \(a=\dfrac{2017}{3}\)

=> \(a=b=c=\dfrac{2017}{3}\)

You are so funny Love the Name of Jiang ha !

After three hours , the length of the car start from A can go :

40 x 3 = 120 (km)

After three hours , the length of the car start from B can go :

45 x 3 = 135 (km)

The length of the street is :

120 + 135 = 255 (km)

x + x² + x³ + ..... + x¹⁰⁰ with x = 1

=> 1 + 1² + 1³ + .... + 1¹⁰⁰

=> 1 + 1 + 1 + .... + 1 (100 number)

=> 1 x 100 = 100 

=> x + x² + x³ + ..... + x¹⁰⁰ = 100

Call the long bottom is a

              short bottom is b 

We have :

\(S=\left(\dfrac{a+b}{2}\right).9=86.4cm^2\) and b = \(\dfrac{1}{2}a\)

Change b to S , we have :

\(\dfrac{a+\dfrac{1}{2}a}{2}.9=86.4\)

\(\dfrac{\dfrac{3}{2}a}{2}=9.6\)

\(\dfrac{3}{2}a.\dfrac{1}{2}=\dfrac{3}{4}a=9.6\)

\(\Rightarrow a=12.8\)

=> b = 6.4

\(\dfrac{a}{b}=\dfrac{a\left(b+c\right)}{b\left(b+c\right)}=\dfrac{ab+ac}{b^2+bc}\)

\(\dfrac{a+c}{b+c}=\dfrac{b\left(a+c\right)}{b\left(b+c\right)}=\dfrac{ab+bc}{b^2+bc}\)

In here , we have :

b² + bc = b² + bc

ab = ab

If a>b

=> ac > bc (when a,b \(\in N\))

=> \(\dfrac{a}{b}>\dfrac{a+c}{b+c}\)

If a<b

=> ac < bc (when a,b \(\in N\))

=> \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

There aren't enough condition !

Give A = 2² + 2⁴ + .... + 2¹⁰⁰

4A = 2⁴ + .... + 2¹⁰⁰ + 2¹⁰²

4A - A = (2⁴ + .... + 2¹⁰⁰ + 2¹⁰²) - (2² + 2⁴ + .... + 2¹⁰⁰)

3A = 2¹⁰² - 2²

A = \(\dfrac{2^{102}-4}{3}\)