the product of roots of equation x(x-2016)(x+2017)=0
Những câu hỏi liên quan
The product of roots of the equation :
X(X-2016)(X+2017)=0
\(x\left(x-2016\right)\left(x+2017\right)=0\)
\(\Rightarrow\)x=0 hoặc x-2016=0 hoặc x+2017=0
\(\Rightarrow\)x=0 hoặc x=2016 hoặc x=-2017
Đúng 0
Bình luận (0)
the number of interger roots of the equation /x-2/+/x-14/=12
/ là giá trị tuyệt đối
the sum of roots of the equation /x-1/=/5-2x/
"/" là giá trị tuyệt đối
Có 2 trường hợp
x-1=5-2x \(\Rightarrow\)x=2
-x+1=-5+2x\(\Rightarrow\)x=-2
Tổng các nghiệm trên = 0
Đúng 0
Bình luận (0)
Khoa Trần trả lời sai rồi đáp án là 6 mới đúng
Đúng 0
Bình luận (0)
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 .
Given the equation \(\frac{3}{x-3}\)- \(\frac{5}{x-5}\) = \(\frac{4}{x-4}\)- \(\frac{6}{x-6}\). The average (arithmetic mean ) of all roots of this equation is.....
\(x\ne\left\{3;4;5;6\right\}\)
\(\frac{3}{x-3}-\frac{5}{x-5}=\frac{4}{x-4}-\frac{6}{x-6}\)
\(\Leftrightarrow\frac{3}{x-3}+1-\frac{5}{x-5}-1=\frac{4}{x-4}+1-\frac{6}{x-6}+1\)
\(\Leftrightarrow\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Leftrightarrow x\left(\frac{1}{x-3}+\frac{1}{x-6}-\frac{1}{x-4}-\frac{1}{x-5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}+\frac{1}{x-6}=\frac{1}{x-4}+\frac{1}{x-5}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-4\right)\left(x-5\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-9=0\\\left(x-3\right)\left(x-6\right)=\left(x-4\right)\left(x-5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\18=20\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\overline{x}=\frac{\frac{9}{2}+0}{2}=\frac{9}{4}\)
Đúng 0
Bình luận (0)
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
Đúng 0
Bình luận (0)
thôi bỏ bài này đi cũng được vì chưa tới lúc cần dung phương trình
Đúng 0
Bình luận (0)
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
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) 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+29right)^2}+dfrac{1}{left(x+30right)^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 ....
Đọc tiếp
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!!!
?????????????????????????????????????????????? Are you learning English or Math? I'm sure you are're mistake of English
Đúng 0
Bình luận (2)
