ĐKXĐ: \(\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}}\)<=>\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

Ta có \(C=\left(x-1\right)-\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)

<=>\(C=\left(x-1\right)-\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}\)

<=>\(C=x-1-\left(2\sqrt{x}+1\right)\)

<=>\(C=x-2\sqrt{x}-2\)

<=>\(C=\left(\sqrt{x}-1\right)^2-3\ge-3\)

Vậy GTNN của C là -3. Dấu "=" xảy ra <=> x=1 (tm ĐKXĐ)

\(B=\dfrac{4x^2-2x+1}{x^2}=\dfrac{3x^2+\left(x^2-2x+1\right)}{x^2}=3+\dfrac{\left(x-1\right)^2}{x^2}\ge3\)

\(B_{min}=3\Leftrightarrow x=1\)

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

Đạo hàm đi bạn :D Cho nhanh

\(y=f\left(x\right)=x^4-2x^2\)

\(\Rightarrow f'\left(x\right)=4x^3-4x\)

\(f'\left(x\right)=0\Leftrightarrow4x^3-4x=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)

\(f\left(1\right)=-1;f\left(-2\right)=8;f\left(-1\right)=-1;f\left(0\right)=0\)

\(\Rightarrow y_{min}=-1;"="\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

\(y_{max}=8;"="\Leftrightarrow x=-2\)

Đặt \(x^2=t\left(0\le t\le4\right)\)

\(y=f\left(t\right)=t^2-2t\)

\(minf\left(t\right)=min\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(1\right)=-1\)

\(maxf\left(t\right)=max\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(4\right)=8\)

\(min=-1\Leftrightarrow x=\pm1\)

\(max=8\Leftrightarrow x=-2\)