First, \(x^2\left(x-3\right)\ne0\) and \(x-9\ne0\Leftrightarrow x\ne0,3,9\)

Then we have \(\left\{{}\begin{matrix}x^2>0\\\dfrac{x^2\left(x-3\right)}{x-9}< 0\end{matrix}\right.\Leftrightarrow\dfrac{x-3}{x-9}< 0\)

We also have \(x-9< x-3\Leftrightarrow\left\{{}\begin{matrix}x-3>0\\x-9< 0\end{matrix}\right.\)

\(\Leftrightarrow3< x< 9\)

Suppose there are \(a\) values of y = \(k\) satisfy the equation above.

When k is satisfied the equation, -k is also satisfied the equation, because \(y^4=k^4=\left(-k\right)^4\)

So the sum of all values of y  is \(a.k+a.\left(-k\right)=0\)

Ans : 0. 

Applying the Ceva theorem, we have :

\(\dfrac{AN}{AP}.\dfrac{BP}{BM}.\dfrac{CM}{CN}=1\)

Substituting the values of AN = 4, CN = 3, CM = 2, BM = 1 and AP = 5, we have :

\(\dfrac{4.BP.2}{5.1.3}=1\Leftrightarrow\dfrac{6}{15}BP=1\Leftrightarrow BP=\dfrac{15}{6}=\dfrac{5}{2}=2,5\)

Ans : 2,5.

\(f\left(a\right)=f\left(b\right)\)

\(\Leftrightarrow a^2+a^3+b=b^2+ab^2+b\)

\(\Leftrightarrow a^2\left(a+1\right)=b^2\left(a+1\right)\)

\(\Leftrightarrow a^2\left(a+b\right)-b^2\left(a+1\right)=0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a+1\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)\left(a+1\right)=0\)

a and b are two different real numbers, so a - b \(\ne0\)

\(\Leftrightarrow\left(a+b\right)\left(a+1\right)=0\)

Scenario 1 :

\(a+b=0\) then a = - b

Substituting the value of a = - b, we have :

\(f\left(x\right)=x^2+ax+\left(-a\right)\)

\(f\left(2\right)=4+2a-a=4+a\)

Scenario 2 :

\(a+1=0\Leftrightarrow a=-1\)

Substituting the value of a = -1, we have :

\(f\left(2\right)=4-2+b=b+2\)

From \(a=\sqrt{6+\sqrt{6+\sqrt{6+...}}}\) we have \(\left\{{}\begin{matrix}a>0\\a^2=6+\sqrt{6+\sqrt{6+...}}=6+a\end{matrix}\right.\)

\(\Leftrightarrow a^2-6-a=0\)

\(\Leftrightarrow\left(a^2-2.\dfrac{1}{2}a+\dfrac{1}{4}\right)-6,25=0\)

\(\Leftrightarrow\left(a-\dfrac{1}{2}\right)^2=6,25\)

\(\Leftrightarrow\left[{}\begin{matrix}a-\dfrac{1}{2}=\dfrac{5}{2}\\a-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-2\end{matrix}\right.\)

\(a>0\Leftrightarrow a=3\)

Again, we have b > 0 and \(b^2=9-b\Leftrightarrow b^2+b-9=0\)

\(\Leftrightarrow\left(b+\dfrac{1}{2}\right)^2-9,25=0\)

\(\Leftrightarrow\left(b+\dfrac{1}{2}\right)^2=9,25\)

\(b>0\Leftrightarrow b+\dfrac{1}{2}>0\)

\(\Leftrightarrow b+\dfrac{1}{2}=\dfrac{\sqrt{37}}{2}\)

Then \(b=\dfrac{\sqrt{37}}{2}-\dfrac{1}{2}=\dfrac{\sqrt{37}-1}{2}\)

\(\Rightarrow ab=3.\dfrac{\sqrt{37}-1}{2}=\dfrac{3\sqrt{37}-3}{2}\)

The NUMBER of pairs (m;n) ... 

\(D=-\dfrac{1}{7}\left(9\dfrac{1}{2}-9,75\right):\dfrac{2}{7}+0,625:1\dfrac{2}{3}\)

\(=\left(-1\right).\dfrac{1}{7}:\dfrac{2}{7}.\left(9,5-9,75\right)+\dfrac{5}{8}:\dfrac{5}{3}\)

\(=-\dfrac{1}{2}.\left(-0,25\right)+\dfrac{3}{8}\)

\(=\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{4+3}{8}=\dfrac{7}{8}\)

\(C=1\dfrac{5}{18}:\dfrac{5}{18}\left(\dfrac{1}{15}+1\dfrac{1}{12}\right)\)

\(=\dfrac{23}{18}:\dfrac{5}{18}:\left(\dfrac{1}{15}+\dfrac{13}{12}\right)\)

\(=\dfrac{23}{5}:\dfrac{23}{20}=\dfrac{20}{5}=4\)

\(\left(x+2\right)^2=28^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{28}=2\sqrt{7}\\x+2=-2\sqrt{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{7}-2\\x=-2\sqrt{7}-2\end{matrix}\right.\)

\(B=\dfrac{2\dfrac{5}{6}+\dfrac{4}{9}}{10\dfrac{1}{12}-9\dfrac{1}{2}}\)

\(=\dfrac{\dfrac{17}{6}+\dfrac{4}{9}}{\dfrac{121}{12}-\dfrac{19}{2}}\)

\(=\dfrac{\dfrac{51}{18}+\dfrac{8}{18}}{\dfrac{121}{12}-\dfrac{114}{12}}\)

\(=\dfrac{\dfrac{59}{8}}{\dfrac{7}{12}}\)

\(=\dfrac{59.12}{7.8}=\dfrac{177}{14}\)

Ans : \(B=\dfrac{177}{14}\).

We have the formulas : \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2017}\left(1+2+...+2017\right)\)

\(=1+\dfrac{\dfrac{2.3}{2}}{2}+\dfrac{\dfrac{3.4}{2}}{3}+...+\dfrac{\dfrac{2017.2018}{2}}{2017}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{2018}{2}\)

\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{2018}{2}\)

\(=\left(\dfrac{1}{2}+\dfrac{2}{2}+\dfrac{3}{2}+...+\dfrac{2018}{2}\right)-\dfrac{1}{2}\)

\(=\dfrac{\left(1+2+...+2018\right)-1}{2}\)

\(=\dfrac{\dfrac{2018.2019}{2}-1}{2}=1018585\)

Ans : A = 1018585.

Copy quên xóa nguồn à :))

Trịnh Đức Phát

You're a thief :)

Because the greatest common dividor of m and n is 15, we put:

 \(\left\{{}\begin{matrix}m=15k\\n=15h\end{matrix}\right.\)\(\left(GCD\left(k;h\right)=1\right)\)

\(\Rightarrow3m+2n=45k+30h=225\)

\(\Rightarrow15\left(3k+2h\right)=225\Rightarrow3k+2h=15\)

+ If h = 0 then k = 5; and that result doesn't satisfy GCD(k;h) = 1.

So h > 0; then k is an odd number. 

\(3k< 15\Rightarrow k< 5\Rightarrow k\in\left\{1;3\right\}\)

If k = 1 then h = 6\(\Rightarrow\) m = 15 ; n = 90 \(\Rightarrow mn=15.90=1350\)

If k = 3 then h = 3; that result doesn't satisfy GCD(k;h)=1.

Therefore, the answer is 1350.

Việt Hoàng, you shouldn't copy Sáng's answer. And Sáng, your English is not so good. May be they can't understand what you have written.

Don't worry, we don't recognize him :)

Given a sum :  2 + 4 + 6 + ... + 2k

The number of terms is : \(\dfrac{2k-2}{2}+1=k\) ( terms )

So, the result of this sum is : \(\dfrac{k.\left(2k+2\right)}{2}=k.\left(k+1\right)\)

Example : 2 + 4 + 6 = 2 + 4 + 2.3 ; so k = 3; then 2 + 4 +6 = k.(k+1)=3.4 .

Dear, Nguyễn Xuân Sáng!

Your English isn't very good, if you continue writing "Land A/B/... = Sth", everybody may not understand your answer.

You can write " Put B = 1.2 + 2. 3 +... ". Use "put", okay :)

If p = 2 then p + 8 isn't a prime number.

If p = 3 then p + 8 and p + 16 are prime numbers.

  \(\Rightarrow3^2+\left(3+8\right)^2+\left(3+16\right)^2=491\)

If p > 3 then \(\left[{}\begin{matrix}p=3k+1\\p=3k+2\end{matrix}\right.\)

  + If p = 3k + 1 then p + 8 = 3k + 9=3(k+1) isn't a prime number.

  + If p = 3k + 2 then p + 16 = 3k + 18=3(k+6) isn't a prime number.

Ans: 491.

a + b + c = 72, then there's an even number and two odd numbers.

The only even prime number is 2, so a = 2.

\(b+c=70=3+67=7+61=11+59=17+53=23+47=29+41\)Ans : a = 2 and \(\left(b;c\right)\in\left\{\left(3;67\right);\left(67;3\right);\left(7;61\right);\left(61;7\right);\left(11;59\right);\left(59;11\right);\left(17;53\right);\left(53;17\right);\left(23;47\right);\left(47;23\right);\left(29;41\right);\left(41;29\right)\right\}\)

\(300=2^2.3.5^2\)

The number of integer factors 300 have is \(2.\left[\left(2+1\right)\left(1+1\right)\left(2+1\right)\right]=36\)

Ans : 36.