phương trình delta

là ntn bn

\(P=ax^2+bx+c=a\left(x^2+\frac{b}{a}x\right)+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\)

Đặt \(c-\frac{b^2}{4a}=k.\)Do \(\left(x+\frac{b}{2a}\right)^2\ge0\)nên:

- Nếu a > 0 thì \(a\left(x+\frac{b}{2a}\right)^2\ge0\). Do đó \(P\ge k\)

min P = k khi và chỉ khi \(x=-\frac{b}{2a}\)

- Nếu a < 0 thì \(a\left(x+\frac{b}{2a}\right)^2\le0\). Do đó \(P\le k\)

max P = k khi và chỉ khi \(x=-\frac{b}{2a}\)

Em tham khảo ở đây:

xét các số thực a,b,c (a≠0) sao cho phương trình ax2+bx+c=0 có 2 nghiệm m, n thỏa mãn \(0\le m\le1;0\le m\le1\). tìm GTN... - Hoc24

Max thì đơn giản thôi em:

Do \(0\le m;n\le1\Rightarrow0< 2-mn\le2\)

\(\Rightarrow M=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{m+n+1}=2\)

\(M_{max}=2\) khi \(mn=0\)

bó tay

\(D=ax^2+bx+c=a\left(x^2+\frac{b}{a}x\right)+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\)

Đặt \(c-\frac{b^2}{4a}=k\). Do \(\left(x+\frac{b}{2a}\right)^2\ge0\)nên :

Nếu a < 0 thì \(a\left(x+\frac{b}{2a}\right)^2\le0\). Do đó \(D\le k\)

Max D = k khi và chỉ khi \(x=-\frac{b}{2a}\)

                              hk tốt

0 nên. đi

a. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c>0\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\b^2-4ac< 0\end{matrix}\right.\end{matrix}\right.\)

b. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b^2-4ac< 0\end{matrix}\right.\end{matrix}\right.\)

c. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\b^2-4ac\le0\end{matrix}\right.\end{matrix}\right.\)

d. \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=b=0\\c\le0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b^2-4ac\le0\end{matrix}\right.\end{matrix}\right.\)

1 ) \(B=x^4-2x^3+3x^2-2x+1\)

       \(B=x^2\left(x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}\right)\)

       \(B=x^2\left[\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1\right]\)

         \(B=x^2\left[\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right]\)

       \(B=x^2\left(x+\frac{1}{x}-1\right)^2\)

       \(B=\left[x\left(x+\frac{1}{x}-1\right)\right]^2\)

       \(B=\left(x^2-x+1\right)^2\)

      Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

           \(\Rightarrow B=\left(x^2-x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{2}\)

2 ) \(A=ax^2+bx+c\) 

     \(A=a\left(x^2+\frac{bx}{a}+\frac{c}{a}\right)\)

     \(A=a\left(x^2+2.x.\frac{b}{2a}+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\)

    \(A=a\left[\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right]\)

    \(A=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\ge\frac{4ac-b^2}{4a}\forall x;a;b;c\)

    Dấu : = " xảy ra \(\Leftrightarrow x=-\frac{b}{2a}\)

Chúc bạn học tốt !!!