If p and q are primes and has \(x^2-px+q=0\) distinct positive integral roots, find p and q .
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If p and q are primes and \(x^2-px+p=0\) has distinct positive integral roots, find p and q.
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 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 .
Ta có:
\(\hept{\begin{cases}ab=q\\a+b=p\end{cases}}\)và \(\hept{\begin{cases}cd=s\\c+d=r\end{cases}}\)
\(M=\frac{2\left(abc+bcd+cda+dab\right)}{p^2+q^2+r^2+s^2}=\frac{2\left(qc+sb+sa+qd\right)}{p^2+q^2+r^2+s^2}\)
\(=\frac{2\left(qr+sp\right)}{p^2+q^2+r^2+s^2}\le\frac{2\left(qr+sp\right)}{2\left(qr+sp\right)}=1\)
Với M = 1 thì \(\hept{\begin{cases}q=r\\p=s\end{cases}}\)
Tới đây thì không biết đi sao nữa :D
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thôi bỏ bài này đi cũng được vì chưa tới lúc cần dung phương trình
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The rectangle ABCD is divided into 4 regions whose perimeters are indicated in the figure below,where X,Y,Z Are Distinct positive integers and X>Y .It is known that Z=\(\frac{X+Y}{3}\)and W<6.Find X
The rectangle ABCD is divided into 4 regions whose perimeters are indicated in the figure below,where X,Y,Z are distinct positive integers and X>Y .It is known that Z=\(\frac{Z+Y}{3}\)and W<6.Find X
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: \(max\left(pq\right)=169\left(cm^2\right)\)
Bạn xem hộ mình sai ở đâu k
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câu 2 thì dựa vào đây nhưng chưa đầy đủ đâu bạn làm nốt nhé https://hoc24.vn/hoi-dap/question/197024.html?pos=675443
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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!
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Toán tiếng anh: A rectangle has length pcm and width qcm, where p and q are integer. If p and q satisfy the equation pq+q=13+q2 then the maximum possible area of the rectangle is.........
pq + q = 13 + q2
<=> p = \(\frac{13+q^2-q}{q}\)
\(S=p\times q=\frac{13+q^2-q}{q}\times q=q^2-q+\frac{1}{4}+\frac{51}{4}=\left(q-\frac{1}{2}\right)^2+\frac{51}{4}\ge\frac{51}{4}\)
ĐS: 12,75
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Given the polynomial P= x^2 + ax +b. Find the values of a and b such that b has 2 roots that are 2 and 3.
a= -5 and b = -1
a= -1 and b = -6
a= 1 and b = 1
a= -5 and b = 6
(giải giùm mình nha các bạn)
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