(m2-4).x2 - (m+2).x + 15 = 0
The equation above, where m is a constant, is a first - degree equation in unknown x if
A) \(m\in\varnothing\) B) m=-2
C) \(m=\pm2\) D) m=2
Những câu hỏi liên quan
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 .
In the equation: N x U x (M + B + E + R) = 33, six letters are 6 distinct digits from 0 to 9. What is the greatest possible value of the number NUMBER?
Answer: .............
In the equation: N x U x (M + B + E + R) = 33, six letters are 6 distinct digits from 0 to 9. What is the greatest possible value of the number NUMBER?
Answer: .............
A solution to the equation (x+a)(x+b)(x+c)+5=0 is x=1 where a,b,c are differents integers Find the value of a+b+c
Thay x=1 vào pt, ta có: \(\left(a+1\right)\left(b+1\right)\left(c+1\right)=-5\left(1\right)\)
vì vai trò của a,b,c là như nhau, giả sử:\(a>b>c\Rightarrow a+1>b+1>c+1\left(2\right)\)
vì a,b,c là số nguyên nên a+1,b+1,c+1 cũng là số nguyên (3)
từ (1),(2),(3)\(\Rightarrow\hept{\begin{cases}a+1=5\\b+1=1\\c+1=-1\end{cases}\Leftrightarrow\hept{\begin{cases}a=4\\b=0\\c=-2\end{cases}}}\)
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In the equation: N x U x (M + B + E + R) = 33, six letters are 6 distinct digits from 0 to 9. What is the greatest possible value of the number NUMBER?
Answer: .............
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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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
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Các bạn giải dùm mình bài này với:
In the equation: N x U x (M + B + E + R) = 33, six letters are 6 distinct digits from 0 to 9. What is the greatest possible value of the number NUMBER?
Cảm ơn các bạn nhiều!
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?