\(-3x^2-9x+6\)

\(=-3\left(x^2+3x-2\right)\)

\(=-3\left(x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{17}{4}\right)\)

\(=-3\left(x+\dfrac{3}{2}\right)^2+\dfrac{51}{4}\le\dfrac{51}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

\(A=-3\left(x^2-\dfrac{5}{3}x+\dfrac{1}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{13}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{13}{12}< =\dfrac{13}{12}\)

Dấu = xảy ra khi x=5/6

a:Ta có: \(A=-4x^2+x-1\)

\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

\(=-\left(x^2-3x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

Bạn tham khảo lời giải tại đây:

Tìm GTLN của biểu thức: \(A=\left(\dfrac{x^2}{x^2-3x 2} \dfrac{x^2}{x^2-5x 6}\right):\dfrac{x^4 x^2 1}{x^2-4x 3}\) - Hoc24

Lời giải:

ĐK: $x\neq 1;2;3$

\(A=x^2\left[\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}\right].\frac{(x-1)(x-3)}{x^4+x^2+1}\)

\(=x^2.\frac{x-3+x-1}{(x-1)(x-2)(x-3)}.\frac{(x-1)(x-3)}{x^4+x^2+1}=x^2.\frac{2(x-2)}{(x-1)(x-2)(x-3)}.\frac{(x-1)(x-3)}{x^4+x^2+1}=\frac{2x^2}{x^4+x^2+1}\)

Áp dụng BĐT AM-GM: $x^4+1\geq 2x^2$

$\Rightarrow A\leq \frac{2x^2}{2x^2+x^2}=\frac{2}{3}$

Vậy $A_{\max}=\frac{2}{3}$. Giá trị đạt tại $x^4=1$ hay $x=-1$ (do $x\neq 1$)

1) \(P=-2x^2-12x=-2\left(x^2+6x+9\right)+18=-2\left(x+3\right)^2+18\le18\)

\(maxP=18\Leftrightarrow x=-3\)

2) \(Q=-5x^2+10x=-5\left(x^2-2x+1\right)+5=-5\left(x-1\right)^2+5\le5\)

\(maxQ=5\Leftrightarrow x=1\)

3) \(A=-3x^2+12x-6=-3\left(x^2-4x+4\right)+6=-3\left(x-2\right)^2+6\le6\)

\(maxA=6\Leftrightarrow x=2\)

4) \(B=-2x^2-24x+12=-2\left(x^2+12x+36\right)+84=-2\left(x+6\right)^2+84\le84\)

\(maxB=84\Leftrightarrow x=-6\)

Ta có

\(1-3x-5x^2=\left(-5x^2-\frac{2.\sqrt{5}x.3}{2\sqrt{5}}-\frac{9}{20}\right)+1+\frac{9}{20}\)

\(=\frac{29}{20}-\left(\sqrt{5}x+\frac{3}{2\sqrt{5}}\right)^2\le\frac{29}{20}\)

Vậy GTLN là \(\frac{29}{20}\) dạt được khi \(\sqrt{5}x+\frac{3}{2\sqrt{5}}=0\Leftrightarrow x=\frac{-3}{10}\)

Câu 1:

\(M=x^2-3x+5\)

\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)

\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

            Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

    Vậy Min M = 11/4 khi x=3/2

b)\(N=2x^2+3x\)

\(N=2\left(x^2+\frac{3}{2}x\right)\)

\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)

\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

              Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)

                       Vậy MIn N = -9/8 khi x=-3/4

c)Tự làm nha

Ta có : x² - 3x + 5 

= x² - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)

= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)

Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)

Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)

Câu 2:

a)\(A=-x^2-5x+3\)

\(A=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{37}{4}\)

\(A=\frac{37}{4}-\left(x+\frac{5}{2}\right)^2\le\frac{37}{4}\)

            Dấu = xảy ra khi \(x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)

                      Vậy Max A = 37/4 khi x=-5/2

b)\(B=-2x^2+3x\)

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

\(B=-2\left(x^2-2.\frac{3}{4}+\frac{9}{16}\right)+\frac{9}{8}\)

\(B=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)

         Dấu = xảy ra khi \(x-\frac{3}{4}=0\Rightarrow x=\frac{3}{4}\)

                    Vậy Max B=9/8 khi x=3/4