Let a,b be the roots of equation \(x^2-px+q=0\) and let c,d be the roots of the equation \(x^2-rx+s=0\), where p,q,r,s are some positive real numbers. Suppose that :
\(M=\frac{2\left(abc+bcd+cda+dab\right)}{p^2+q^2+r^2+s^2}\)
is an integer. Determine a,b,c,d .
let a<b<c<d be integers. If one of the roots pf the equation (x-a)(x-b)(x-c)(x-d) - 9 is x=7, what is a+b+c+d?
1) The rectangle has length p and breath q (cm), where p and q are intergers. If p and q satisfy the equation pq+q=13 + q2
then the maxnium area of the rectangle
2) Let a,b and c be positive intergers such that ab + bc=518 and ab-ac=360. Find the largest value of the product abc.
P/s: As you may now, These are some questions from the 8 round of Math Violympic. Plz help me as much as you can! Thanks for all!
Ta có: \(pq+q=13+q^2\Leftrightarrow q\left(p+1\right)=13+q^2\)
Vì\(q^2⋮q\Leftrightarrow13⋮q\Leftrightarrow\left[{}\begin{matrix}q=1\\q=13\end{matrix}\right.\)
Nếu q =1 thì:\(p+1=14\Leftrightarrow p=13\)
\(\Rightarrow pq=13\left(cm^2\right)\)(1)
Nếu q=13 thì:\(13p+13=182\Leftrightarrow p=13\)
\(\Rightarrow pq=169\left(cm^2\right)\)(2)
Từ (1)(2) ta có: \(max\left(pq\right)=169\left(cm^2\right)\)
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Hey guys! I have to do a City's Math Violympic on March 9th ( who like me, raise hands, lol :)) so it is not enough time to solve this very difficult problem, right??? Plz help me,guys! God bless you all xx. Sending you a big hug!
It is known that the roots of the equation \(3x^5+9x^4-6x^2+5x-7=0\)are all integers.How many distinct roots does the equation have
Determine all positve integer a such that the equation \(2x^2-210x+a=0\) has two prime roots, i.e. both roots are prime numbers
Determine all positve integer a such that the equation \(2x^2-210x+a=0\) has two prime roots, i.e. both roots are prime numbers
Phương trình: \(2x^2-105x+a=0\Leftrightarrow x^2-105x+\frac{a}{2}=0\)không thể có nghiệm kép được vì 105 là số lẻ
Giả sử phương trình này có 2 nghiệm là b, c ta có
\(\hept{\begin{cases}2b^2-210b+a=0\left(1\right)\\2c^2-210c+a=0\left(2\right)\end{cases}}\)
Lấy (1) - (2) vế theo vế ta được
\(2b^2-210b-2c^2+210c=0\)
\(\Leftrightarrow\left(b-c\right)\left(b+c-105\right)=0\)
\(\Rightarrow b+c-105=0\Leftrightarrow b+c=105\)
\(\Rightarrow\)Một trong 2 số b hoặc c phải là số chẵn
Giả sử số chẵn đó là c thì ta có c = 2 ( vì c nguyên tố)
\(\Rightarrow b=103\)
Từ đây ta có:\(x^2-105x+\frac{a}{2}=\left(x-2\right)\left(x+103\right)=x^2-105x+206\)
\(\Rightarrow a=2.206=412\)
Let a , b and c be positive real numbers such that a + b + c = 3 . Find the minimum value of the expression .
\(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2+b^2+c^2}\)
1) ABC is a triangle where M is the midpoint of segment BC.
MD and ME are two bisectors of triangles AMB and AMC respectively.
If AM= m; BC = a . Then DE = ???
2)\(\dfrac{1}{\left(x+29\right)^2}+\dfrac{1}{\left(x+30\right)^2}=\dfrac{5}{4}\)
What is the product of all real solutions to the equation above?
3) The sum of all possible natural numbers n such that
\(n^2+n+1589\) is a perfect square is.....
4) Given that x is a positive integer such that x and x+99 are perfect squares
The sum of integer x is ...
5)The operation @ on two numbers produces a number equal to their sum minus 2. The value of
(...((1@2)@3....@2017)
6) Given f(x)=\(\dfrac{x^2}{2x-2x^2-1}\)
=> \(f\left(\dfrac{1}{2016}\right)+f\left(\dfrac{2}{2016}\right)+f\left(\dfrac{3}{2016}\right)+...+f\left(\dfrac{2016}{2016}\right)\)
Các bn giúp mk vs >>> tks nha!!!
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the product of roots of equation x(x-2016)(x+2017)=0
