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The expression is determined with all \(x\in R\)

\(\Leftrightarrow3\left(x^2+x+3\right)-\sqrt{x^2+x+3}-24=0\)

Let \(\sqrt{x^2+x+3}=a>0\Rightarrow3a^2-a-24=0\Leftrightarrow\left(a-3\right)\left(3a+8\right)=0\)

\(\Leftrightarrow a=3\left(cause:3a+8>0\right)\)

So  the expression is equal : \(\sqrt{x^2+x+3}=3\)

\(\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

So the expression has two solutions : x = 2 ; x = -3.

The degree measure of the largest angle of the triangle is \(78^o\)

The triple measure angles are : \(42^o;60^o;78^o\)

This problem is easy we start : 

Note : \(3x^2+6x+7=3\left(x^2+2x+1\right)+4=3\left(x+1\right)^2+4\Rightarrow\sqrt{3x^2+6x+7}\ge2\)

\(5x^2+10x+14=5\left(x^2+2x+1\right)+9=5\left(x+1\right)^2+9\Rightarrow\sqrt{5x^2+10x+14}\ge3\)

So the left side of the expression \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}\ge5\)

Come to the right side : \(4-2x-x^2=5-\left(x^2+2x+1\right)=5-\left(x+1\right)^2\le5\)

The thread \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=5\\4-2x-x^2=5\end{matrix}\right.\Leftrightarrow x=-1\)

So the expression has the only solution x = -1.

Condition : \(0\le x\le1;x\ne\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\left(6x-3\right)\left(\sqrt{x}+\sqrt{1-x}\right)}{2x-1}=3+2\sqrt{x-x^2}\)

\(\Leftrightarrow3\cdot\left(\sqrt{x}+\sqrt{1-x}\right)=x+2\sqrt{x\left(1-x\right)}+\left(1-x\right)+2\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{1-x}\right)^2-3\left(\sqrt{x}+\sqrt{1-x}\right)+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x\left(1-x\right)}=0\\4x^2-4x+9=0\left(no-solution\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) So the expression has two solutions : x = 0; x = 1.

The expression is determined with all \(x\in R\)

\(\Leftrightarrow x^2+1+3x-x\sqrt{x^2+1}-3\sqrt{x^2+1}=0\)

\(\Leftrightarrow\left(x^2+1\right)^2-x\sqrt{x^2+1}+3x-3\sqrt{x^2+1}=0\)

\(\Leftrightarrow\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)-3\left(\sqrt{x^2+1}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(not-satisfy\right)\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow x=\pm2\sqrt{2}\)

1) \(x^3-25x=0\)

\(\Leftrightarrow x\cdot\left(x^2-25\right)=0\)

\(\Rightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

2) \(2x\left(x-1\right)-3x+3=0\)

\(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=1\end{matrix}\right.\)

3) \(x^2-36+\left(x^2-12x+36\right)=0\)

\(\Leftrightarrow x^2-36+\left(x-6\right)^2=0\)

\(\Leftrightarrow\left(x-6\right)\cdot\left(x+6\right)+\left(x-6\right)^2=0\)

\(\Leftrightarrow\left(x-6\right)\cdot\left[\left(x+6\right)+\left(x-6\right)\right]=0\)

\(\Leftrightarrow\left(x-6\right)\cdot2x=0\)

\(\Rightarrow\left[{}\begin{matrix}x-6=0\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

I have another way more logical than Phan Thanh Tinh's answer ! :)

Here it is : 

\(x^2+2005x+2006y^2+y=xy+2006xy^2+2007\)

\(\Leftrightarrow\left(x^2+2005x-2006\right)+\left(2006y^2-2006xy^2\right)+\left(y-xy\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+2006\right)-2006y^2\left(x-1\right)-y\left(x-1\right)=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+2006-2006y^2-y\right)=1\) (1)

Cause x,y are interger numbers so 2 factors on the left side of expression (1) is divisor of 1.

So it occurs two cases :

Case 1 : \(\left\{{}\begin{matrix}x-1=1\left(2\right)\\x+2006-2006y^2-y=1\left(3\right)\end{matrix}\right.\)

From (2) -> x = 2.

(3) \(\Leftrightarrow2006y^2+y-2007=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-\dfrac{2007}{2006}\notin Z\end{matrix}\right.\)

So (x;y) = (2;1).

Case 2 : \(\left\{{}\begin{matrix}x-1=-1\left(4\right)\\x+2006-2006y^2-y=-1\left(5\right)\end{matrix}\right.\)

From (4) => x = 0.

(5) \(\Leftrightarrow2006y^2+y-2007=0\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-\dfrac{2007}{2006}\notin Z\end{matrix}\right.\)

So (x;y) = (0;1).

So the expression has two pairs of interger numbers (x;y) are (2;1) and (0;1).

This is a good and hard problem, I have a way to solve this but it's not mine :)

Here it is : 

We know that  (where , , and  are the hundreds, tens, and ones digit respectively)

Case 1: 

We can use the kids and candy method (a.k.a. stars and bars, ss and stones etc.) method which is denoted by  where  stands for the amount of kids (in this case, its the  variables) and  stands for
the number of candies. (in this case, it's the overall sum, which is ) Substituting, we get


 ways to have  numbers sum up to . We can use the kids and candy technique in the other cases. We just have to substitute the correct values.

Case 2: 

 ways for  numbers to sum up to .

Case 3: 

 ways of  numbers to sum up to .

Case 4: 

 ways of  numbers to sum up to .

Case 5: 

 ways of  numbers to sum up to .

Now we just add up all the cases to find the answer as  numbers that meet the criteria.

We have : \(LCM\left(8;5;10\right)=40\)

So he will take the medicines together forty hours afterwards or one day and 16 hours.

That will be at 11 p.m on Wednesday.

Okay :0

\(999999\cdot222222=2,222217778\times_{10^{11}}\)

Minus \(2\times10^{11}\) we have \(2,222177778\times_{10^{10}}\)

Minus \(2\times10^{10}=221777778\)

So \(999999\cdot222222=222221777778\)

Next : \(333333\cdot333334=1,111112222\times_{10^{11}}\)

Do as upper we have answer : \(333333\cdot333334=111111222222\)

So \(999999\cdot222222+333333\cdot333334=222221777778+111111222222=333333000000\)

From A to B, we can see that the way go by side (not go inside) is equal :

14+6+17+12=15+7+9+18 (=49km).

From A to midpoint we have two way are : 15 + 11 (=26km) and 14+13(=27km)

From midpoint to B we have two way are 10 + 12 (= 22km) and 5+18(=23km).

So the shortest way from A to midpoint to B is 15+11+10+12 (=48km).

Compare the way from A to midpoint to B with the way go by side : 48 < 49.

So the shortest distance from A to B is 48 km.

Call the price of large is x, the price of small is y.

We have : \(\left\{{}\begin{matrix}7x+4y=100\\5x+6y+5=100\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x+4y=100\\5x+6y=95\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\y=7,5\end{matrix}\right.\)

So a small exercise book costs 7,5$.

\(T=\dfrac{1}{1\cdot2}+\dfrac{5}{2\cdot3}+\dfrac{11}{3\cdot4}+...+\dfrac{89}{9\cdot10}\)

\(T=\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{9\cdot10}\right)\)

\(T=9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\)

\(T=9-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(T=9-\left(\dfrac{1}{1}-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)

Call the number is n.

We have : \(n=342q+47\)

\(\Leftrightarrow n=19\cdot18q+19\cdot2+9=19\cdot\left(18q+2\right)+9\)

So when divide 19 the remainder would be 9.

Call the radius of center circle A is AO, B is BI.

We have AO + BI = 2 + 2 = 4 (cm).

<=> AO + BI - OI = AB = 4 - (2:2) = 4 - 1 = 3 (cm).

That's my idea :((

Okay :0

\(\dfrac{501}{2015}=\dfrac{1}{\dfrac{2015}{501}}=\dfrac{1}{4+\dfrac{11}{501}}=\dfrac{1}{4+\dfrac{1}{\dfrac{501}{11}}}=\dfrac{1}{4+\dfrac{1}{45+\dfrac{6}{11}}}=\dfrac{1}{4+\dfrac{1}{45+\dfrac{1}{\dfrac{11}{6}}}}=\dfrac{1}{4+\dfrac{1}{45+\dfrac{1}{1+\dfrac{5}{6}}}}=\dfrac{1}{4+\dfrac{1}{45+\dfrac{1}{1+\dfrac{1}{\dfrac{6}{5}}}}}=\dfrac{1}{4+\dfrac{1}{45+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{5}}}}}\)

So (a;b;c;d;e) = (4;45;1;1;5)

Let \(\dfrac{X+Y}{X-Y}_{MAX}\) so X-Y must in range 1 - the minimum range (X-Y mustn't in range 0 because it is denominator).

So from 1 to 40 we can choose \(\left\{{}\begin{matrix}X=21\\Y=20\end{matrix}\right.\Leftrightarrow\dfrac{X+Y}{X-Y}=\dfrac{21+20}{21-20}=\dfrac{41}{1}=41\)

So the largest value that \(\dfrac{X+Y}{X-Y}\) can have is 41.

Cause \(\overline{3AA1}⋮9\)

So \(\left(3+2A+1\right)⋮9\Leftrightarrow\left(4+2A\right)⋮9\)

And \(0\le A\le9\) (Cause A is a number which can be represented in a four-digit number).

Make a table and do calculating we have A represent the number 7 which \(\left(4+2\cdot7\right)=18⋮9\)

So the digit that A represents is 7.

Power of 2, 2 in alone version ?

\(P=4+2^2+2^3+2^4+....+2^{20}\)

\(2P=8+2^3+2^4+2^5+...+2^{21}\)

\(\Rightarrow2P-P=P=\left(8+2^3+2^4+2^5+...+2^{21}\right)-\left(4+2^2+2^3+2^4+...+2^{20}\right)=\left(8+2^{21}\right)-\left(4+2^2\right)=8+2^{21}-8=2^{21}\)

So P can be be written as a power of 2 and write  \(P=2^{21}\)