When type in calculator we have :

\(13\cdot53^2\cdot3853\cdot96179=1.35323854\times10^{13}\)

Minus \(1\times10^{13}\) from that result we have : \(3.532385396\times10^{12}\)

Minus \(3\times10^{12}=5.323853962\times10^{11}\)

Minus \(5\times10^{11}-3\times10^{10}=2385396179\)

Correct answer : \(13532385396179\)

You are right.

In an isosceles triangle, if all the angles are equal => We have equilateral triangle.

So all the sides of equilateral triangle are equal. 

Call one side of equilateral triangle a. => We have an median drawn from opposite vertice down to a/2.

So the same to two others, we also drawn from opposite vertice down to a/2.

So in an isosceles triangle, medians drawn from vertices with equal angles are equal in length.

We have the theorem : In the isosceles triangle, the height is also the midperpendicular and also the median.

So in ABC isosceles triangle ABC, we have \(H\in BC\) so AH is the median of ABC isosceles triangle.

=> BH = BC.

Next, we have \(D\in BC\) satisfy \(DB=DC\)(1)

But BH also equal BC (2).

(1),(2) => \(H\equiv D\)

It's a good problem :((

So we have the time is \(\dfrac{2}{60}\left(hour\right)=\dfrac{1}{30}\)

So we have the speed to get to each intersection right on the first time is : \(\dfrac{3,5}{x}=\dfrac{1}{30}\Rightarrow x=105\) (kilometres per hour).

So he can travel to get to each intersection at the third time as it just changes to green at the speed of \(\dfrac{105}{3}=35\) (kilometres per hour)

That's what I think :))

The answer is 901.

From 1 to 9 we have sum : 45.

From 10 to 19 we have : 1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 +...+ 1 + 9 we have sum 55.

Next from 20 to 29 we have sum 65.

................. 30 to 39 we have sum 75.

................. 40 to 49 we have sum 85.

................. 50 to 59 we have sum 95.

................. 60 to 69 we have sum 105.

................. 70 to 79 we have sum 115.

................. 80 to 89 we have sum 125.

................. 90 to 99 we have sum 135.

And the last 100 we have sum 1 + 0 + 0 = 1.

So the sum she obtained is : 65 + 75 + 85 + 95 +105 +115 +125 +135 + 1 = 901.

\(\dfrac{x^2-3x+2}{x^2+4x+5}=\dfrac{1}{2}\Rightarrow2\left(x^2-3x+2\right)=x^2+4x+5\)

\(\Leftrightarrow2x^2-6x+4=x^2+4x+5\)

\(\Leftrightarrow x^2-10x-1=0\)

=> We have two solutions : \(\left[{}\begin{matrix}x=5+\sqrt{26}\\x=5-\sqrt{26}\end{matrix}\right.\)

Use Cauchy's inequality for positive numbes a,b,c.

\(a+b\ge2\sqrt{ab},b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab}\cdot\sqrt{bc}\cdot\sqrt{ca}=8abc\)

\(\Rightarrow A=\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)

\(MaxA=\dfrac{1}{8}\Leftrightarrow a=b=c>0\)

Real Egst ?

a) \(x-x^2+x^3-x^4\)

\(\Leftrightarrow x\cdot\left(1-x\right)+x^3\cdot\left(1-x\right)\)

\(\Leftrightarrow\left(x+x^3\right)\cdot\left(1-x\right)\)

b) \(x^2+2x+3x+x\)

\(\Leftrightarrow x\cdot\left(x+6\right)\)

Hiệp Dương : You answer very fast, sir :))

More than 8 lines but you just answer in one minute.

Probably copier :))

Just "Calculate" not "Caculader" :) Wrong vocabulary sir :D

x(x + 3) - x - 3 = 0

\(\Leftrightarrow x^2+3x-x-3=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

So, \(S=\left\{-1;3\right\}\)

2x - x + 3x - 2x = 0 

\(\Leftrightarrow x+x=0\)

\(\Rightarrow x=-x\)

We just have one answer for this math problem.

That's \(x=0\)

So, \(S=\left\{0\right\}\)

When you meet high variable  \(x^4,x^5...\) the way for expressing this is change it to factored equations (trans : phương trình tích).

Now change that equation into : \(\left(x+y\right)\left(x-y\right)^2=5\)

So the equation has interger solutions (x;y) = (3;2) , (2;3).

Good luck.

This problem has a lot of ways to express.

\(S=1-2+3-4+...+2007-2008+2009-2010\)

We mark each pair to each other until the answer of each pair is -1 :

\(S=\left(1-2\right)+\left(3-4\right)+...+\left(2007-2008\right)+\left(2009-2010\right)\)

From here we have the number of pairs are : \(\dfrac{\left(\dfrac{2010-1}{1}+1\right)}{2}=1005\)(pairs)

Then the correct answer is : \(-1\cdot1005=-1005\)

Both answers upper there are correct, I just want to donate one more way to express this math prob :)

You're all copiers, your both answers don't have any differences.

You shouldn't be in this website :))

First, call I the intersection of AM and AN.

We have : \(4BN^2=4AB^2+AC^2\)

\(AI^2=AB^2-BI^2=AB^2-\dfrac{4}{9}BN^2\Rightarrow\dfrac{4}{9}AM^2=AB^2-\dfrac{4}{9}\left(AB^2+\dfrac{1}{4}AC^2\right)\)

\(\Rightarrow\dfrac{1}{9}BC^2=\dfrac{5}{9}AB^2-\dfrac{1}{9}AC^2\)

\(\Rightarrow BC^2=5AB^2-AC^2\) and \(BC^2=AB^2+AC^2\Rightarrow AC^2=2AB^2\)

From here we can know that AC = 4.

Good Luck !

2. Remark : \(1+a^2=ab+bc+ca+a^2=\left(a+b\right)\left(a+c\right)\)

\(1+b^2=ab+bc+ca+b^2=\left(b+a\right)\left(b+c\right)\)

\(1+c^2=ab+bc+ca+c^2=\left(c+a\right)\left(c+b\right)\)

Using Cauchy for these positive numbers \(\dfrac{1}{a+b},\dfrac{1}{b+c},\dfrac{1}{c+a}\)

We have : \(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{2}{\sqrt{\left(a+b\right)\left(b+c\right)}}\Rightarrow\dfrac{b}{a+b}+\dfrac{b}{b+c}\ge\dfrac{2b}{\sqrt{1+b^2}}\)

\(\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{2}{\sqrt{\left(b+c\right)\left(c+a\right)}}\Rightarrow\dfrac{c}{b+c}+\dfrac{c}{c+a}\ge\dfrac{2c}{\sqrt{1+c^2}}\)

\(\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{2}{\sqrt{\left(c+a\right)\left(a+b\right)}}\Rightarrow\dfrac{a}{c+a}+\dfrac{a}{a+b}\ge\dfrac{2a}{\sqrt{1+a^2}}\)

Plus each pair of inequalities we have :

\(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\ge2\left(\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\right)\)

\(\Rightarrow\dfrac{3}{2}\ge\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)

Yes, thanks Egst for your answers.

I also have another ideas for these math problems :

1. Using Bunhiacopxki, we have :

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

\(\Rightarrow2\left(a^2+b^2\right)\ge4\Rightarrow a^2+b^2\ge2\)

We have : \(\left(a^2-1\right)^2\ge0\Rightarrow a^4+1\ge2a^2\)

\(\left(b^2-1\right)^2\ge0\Rightarrow b^4+1\ge2b^2\)

Plus each pair of the two inequalities we have :

\(a^4+b^4+2\ge2\left(a^2+b^2\right)\)

\(\Rightarrow a^4+b^4\ge a^2+b^2+\left(a^2+b^2-2\right)\ge a^2+b^2\) because \(a^2+b^2\ge2\)

So \(a^4+b^4\ge a^2+b^2\)

Anyone who has different answer plz comment in this question ;D

The solution of (a;b) is (1;2) and (2;1) => These two solutions are equal.

When (a;b) equal (1;2) => \(a^5+b^5=1^5+2^5=33\)

When (a;b) equal (2;1) => \(a^5+b^5=2^5+1^5=33\)

Need to block the copy mode in this website.

Suggest the moderator to apply the deleting mode for each Coordinator to help this better.

Create one more team (have badges) in order to help this probs low down.

That's my idea about this prob/