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ForOver Coordinator 13/10/2017 at 11:44

We have \(\overline{a0b}⋮9\Rightarrow a+0+b\in\left\{9;18;27;36;...\right\}\)

But \(a+0+b\le18\)

So \(a+b=\left\{9;18\right\}\)

We have pair (a;b) = (1;8) , (8;1) , (2;7) , (7;2) , (6;3) , (3;6) , (5;4) , (4;5) , (9;0).

But there is only the pair : (4;5) match the condition \(\overline{a0b}=\overline{ab}\cdot9\).

So the answer of the two digit number \(\overline{ab}\) is 45.
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Phan Thanh Tinh Coordinator 13/10/2017 at 11:48

\(\overline{a0b}=\overline{ab}.9\Leftrightarrow100a+b=9\left(10a+b\right)\Leftrightarrow100a+b=90a+9b\)

\(\Leftrightarrow10a=8b\Leftrightarrow5a=4b\Leftrightarrow\dfrac{a}{4}=\dfrac{b}{5}\)

\(b< 10\Rightarrow\dfrac{b}{5}< 2\). Moreover, \(a>0\Rightarrow\dfrac{a}{4}>0\)

So, \(\dfrac{a}{4}=\dfrac{b}{5}=1\Rightarrow a=4;b=5;\overline{ab}=45\)

Vũ Mạnh Hùng 02/12/2017 at 20:50

it is 36 letters me
Shirayuki Akagami 02/12/2017 at 13:08

I do not know ???

Alone 07/04/2018 at 08:27

\(\left(2x_1-5y_1\right)^{2018}\ge0;\left(2x_2-5y_2\right)^{2018}\ge0;.......;\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)

\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)

Because \(\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+..........+\left(2x_{2019}-5y_{2019}\right)^{2018}\le0\)

\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}=0\)

\(\Rightarrow2x_1=5y_1;2x_2=5y_2;..........;2x_{2019}=5y_{2019}\)

\(\Rightarrow\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=..........=\dfrac{x_{2019}}{y_{2019}}=\dfrac{5}{2}=\dfrac{x_1+x_2+.......+x_{2019}}{y_1+y_2+.......+y_{2019}}\)
Dao Trong Luan selected this answer.
¤« 07/04/2018 at 14:38

(2x1−5y1)2018≥0;(2x2−5y2)2018≥0;.......;(2x2019−5y2019)2018≥0

⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018≥0

Because (2x1−5y1)2018+(2x2−5y2)2018+..........+(2x2019−5y2019)2018≤0

⇒(2x1−5y1)2018+(2x2−5y2)2018+......+(2x2019−5y2019)2018=0

⇒2x1=5y1;2x2=5y2;..........;2x2019=5y2019

⇒x1y1=x2y2=..........=x2019y2019=52=x1+x2+.......+x2019y1+y2+.......+y2019

Cristiano Ronaldo 03/08/2018 at 14:03

How your father feel???

Answer: he happy too
Hello selected this answer.

Trần Nhật Dương 26/06/2017 at 15:32

Draw AE // BC.

The quadrilateral ABCE has AB // CE ; AE // BC,so it's a parallelogram

$\Rightarrow A B = C E; \hat{E_{1}} = \hat{B} = 125^{0} \Rightarrow \hat{E_{2}} = 180^{0} - \hat{E_{1}} = 55^{0}$

$\hat{E_{1}}$ is the exterior angle of $Δ A D E$ $\Rightarrow \hat{A_{1}} + \hat{D} = \hat{E_{1}}$

$\Rightarrow \hat{A_{1}} = 125^{0} - 70^{0} = 55^{0}$

$Δ A D E$ has $\hat{A_{1}} = \hat{E_{2}} = 55^{0}$ ,so $Δ A D E$ isosceles at D

$\Rightarrow A D = D E \Rightarrow C D = C E + D E = A B + A D$

P/S : The question must be changed into $\hat{D} = 70^{0}$ or $\hat{B} = {127.5}^{0}$
Phan Thanh Tinh Coordinator 26/06/2017 at 10:57

Draw AE // BC.

The quadrilateral ABCE has AB // CE ; AE // BC,so it's a parallelogram

\(\Rightarrow AB=CE;\widehat{E_1}=\widehat{B}=125^0\Rightarrow\widehat{E_2}=180^0-\widehat{E_1}=55^0\)

\(\widehat{E_1}\)is the exterior angle of \(\Delta ADE\) \(\Rightarrow\widehat{A_1}+\widehat{D}=\widehat{E_1}\)

\(\Rightarrow\widehat{A_1}=125^0-70^0=55^0\)

\(\Delta ADE\) has \(\widehat{A_1}=\widehat{E_2}=55^0\),so \(\Delta ADE\) isosceles at D

\(\Rightarrow AD=DE\Rightarrow CD=CE+DE=AB+AD\)

P/S : The question must be changed into \(\widehat{D}=70^0\)or \(\widehat{B}=127.5^0\)

Kaya Renger Coordinator 16/08/2017 at 09:00

\(1-5\cdot2+10\cdot2^2-10\cdot2^3+5\cdot2^4-2^5\)

\(=1-5\cdot2+5\cdot2^3-5\cdot2^4+5\cdot2^4-2^5\)

\(=1-5\cdot2+5\cdot2^3-2^5\)

\(=1-5\cdot\left(2-2^3\right)-2^5\)

\(=1-5.\left(-6\right)-32\)

\(=1+30-32=-1\)
Lê Quốc Trần Anh selected this answer.
Help you solve math 16/08/2017 at 15:36

Answer:1−5⋅2+10⋅22−10⋅23+5⋅24−251−5⋅2+10⋅22−10⋅23+5⋅24−25

=1−5⋅2+5⋅23−5⋅24+5⋅24−25=1−5⋅2+5⋅23−5⋅24+5⋅24−25

=1−5⋅2+5⋅23−25=1−5⋅2+5⋅23−25

=1−5⋅(2−23)−25=1−5⋅(2−23)−25

=1−5.(−6)−32=1−5.(−6)−32

=1+30−32=−1

Recipe: