We have :

AH . BC = AB . AC (equals two \(S_{\Delta ABC}\))

So 2.AH.BC = 2.AB.AC  (1)

Apply Pythagoras theorem , we have :

AB² + AC² = BC²  (2)

But (AH + BC)² = AH² + BC² + 2.AH.BC   (3)

(AB + AC)² = AB² + AC² + 2AB.AC   (4)

From (1) ; (2) ; (3) ; (4) 

=> AH + BC > AB + AC

Câu hỏi của hung - Toán lớp 8 - Học toán với OnlineMath

Call the formular of numbers we need to find is \(\overline{abcd}\)

Condition : m + k < 200 

Following the thread, we have :

\(\left\{{}\begin{matrix}\overline{abcd}=k^2\\\overline{\left(a+1\right)\left(b+3\right)\left(c+5\right)\left(d+3\right)}=m^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\overline{abcd}=k^2\\\overline{abcd}+1353=m^2\end{matrix}\right.\)

=> m² - k² = 1353 

<=> (m - k)(m + k) = 1353 = 123.11 = 41.33 

=> \(\left\{{}\begin{matrix}m+k=123\\m-k=11\end{matrix}\right.\)  or  \(\left\{{}\begin{matrix}m+k=41\\m-k=33\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m=67\\k=56\end{matrix}\right.\)    or    \(\left\{{}\begin{matrix}m=37\\k=4\end{matrix}\right.\)

With m = 67 ; k = 56

=> \(\left\{{}\begin{matrix}\overline{abcd}=k^2=3136\\\overline{abcd}=m^2-1353=3136\end{matrix}\right.\Rightarrow\overline{abcd}=3136\)

With m = 37 , k = 4

=> .......... (This case is disqualified , you can solve that equation)

So , the number we need to find is 3136 

I need another way , not that way ! 

Denote A = (x + 1)(x + 3)(x + 5)(x + 7) + 2013

            A = (x² + 8x + 7)(x² + 8x + 15) + 2013

            A = (x² + 8x + 12 - 5)(x² + 8x + 12 + 3) + 2013

Let t = x² + 8x + 12

A = (t - 5)(t + 3) + 2013

A = t² + 3t - 5t - 15 + 2013

A = t(t - 2) + 1998

Because \(t\left(t-2\right)⋮x^2+8x+12\)

=> \(\dfrac{A}{x^2+8x+12}\left(surplus1998\right)\)

I had did it for a few minutes ago :v 

a) (x - 3)³ - (x - 3)(x² + 3x + 9) + 6(x + 1)² = 15

(x - 3)*[(x - 3)² - (x² + 3x + 9)] + 6(x² + 2x + 1) = 15

(x - 3)*[x² - 6x + 9 - x² - 3x - 9] + 6x² + 12x + 6 = 15

(x - 3)*(-9x) + 6x² + 12x + 6 = 15

-9x² + 27x + 6x² + 12x + 6 = 15

39x - 3x² = 9

13x - x² = 3

x² - 13x = -3

x² - 13x + 3 = 0

\(x^2-2.x.\dfrac{13}{2}+\dfrac{169}{4}-\dfrac{157}{4}=0\)

\(\left(x-\dfrac{13}{2}\right)^2=\dfrac{157}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{2}=\dfrac{\sqrt{157}}{2}\\x-\dfrac{13}{2}=\dfrac{-\sqrt{157}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13+\sqrt{157}}{2}\\x=\dfrac{13-\sqrt{157}}{2}\end{matrix}\right.\)

b) x(x - 5)(x + 5) - (x + 2)(x² - 2x + 4) = 3

x(x² - 25) - (x³ + 8) = 3

x³ - 25x - x³ - 8 = 3

-25x = 11

\(x=-\dfrac{11}{25}\)

\(f\left(10\right)=3\cdot10-4=30-4=26\)

\(g\left(-4\right)=\left(-4\right)^2-2.\left(-4\right)+5=16+8+5=29\)

\(\Rightarrow g\left(-4\right)-f\left(10\right)=29-26=3\)

\(-\dfrac{45}{47}>-1\)

\(\dfrac{31}{-30}\Leftrightarrow\dfrac{-31}{30}< -1\)

=> \(\dfrac{31}{-30}< \dfrac{-45}{47}\)

First note that there are only 2 ways in which Abby can sit: the left-most seat or the right-most seat. Say Abby sits on the left. Then since Bea and Jaclyn sit together, there are 4 ways to seat them. We then multiply this by 2 since Abby could also sit on the right side. Therefore there are 8 ways to seat them.

With n is the number of edges that polygon have , we have this formula to calculate the sum of interior angles :

\(\left(n-2\right).180^0\)

So , apply this formula into the thread , we have : 

The sum of the degree measures of the interior angles of a regular octagon is :

\(\left(8-2\right)\cdot180^0=1080^0\)

The number of seconds does Candace run ahead Vicky is :

29'46'' - 28'47'' = 0'59'' <=> 59 seconds

So ...... 

Call the point of sixth quiz is a

Following the thread , we have :

\(\dfrac{80+84+99+92+100+a}{6}=90\)

<=> \(\dfrac{455+a}{6}=90\)

<=> 455 + a = 540

<=> a = 540 - 455

=> a = 85

10 + 20 + 30 + 40 + 50 + 60 = (1 + 2 + 3 + 4 + 5 + 6) * 10

                                              = 21 * 10

                                              = 210 

Call three consecutive natural numbers are n; n + 1 ; n + 2  (n \(\in\) N)

If n divide 3 balance 0 => n = 3q (q \(\in\) N)

=> n = 3q \(⋮\) 3

If n divide 3 balance 1 => n = 3q + 1 (q \(\in\) N)

=> n + 2 = (3q + 1) + 2 = 3q + 3 = 3(q + 1) \(⋮\) 3

If n divide 3 balance 2 => n = 3q + 2 (q \(\in\) N)

=> n + 1 = (3q + 2) + 1 = 3q + 3 = 3(q + 1) \(⋮\) 3

So , with three consecutive natural numbers , there is always exist a number divisible by 3 .

The perimeter of the recangle is :

\(P=\left(12+8\right)\cdot2=20\cdot2=40\left(cm\right)\)

Câu hỏi của Phan Huy Tiến - Toán lớp 8 - Học toán với OnlineMath

Phanhuytien , I do not need your answer anymore ! 

A is the midpoint of BD => BA = BD

Triangle ABC isosceles at A => AB = AC

=> AB = AD = AC 

But BD is the hypothenuse side 

=> triangle BCD right at C

We have got 

\(\widehat{FCE}=\widehat{AFC\:}=\widehat{AEC}=90^0\)

=> Quadrilateral AFCE is the rectangle

=> \(\widehat{EAF}=90^0\)

a) \(S_1=1+2+3+....+999=\dfrac{\left[\left(999-1\right):1+1\right].\left(999+1\right)}{2}=999.1000:2=999.500=499500\)b) \(S_2=10+12+14+...+2010\dfrac{\left[\left(2010-10\right):2+1\right].\left(2010+10\right)}{2}=1001.2020:2=1001.1010=1011010\)

D) 5 Thursday 

\(M=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(M=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(M=\dfrac{5}{7}.\dfrac{-7}{11}=-\dfrac{5}{11}\)