Tìm GTNN của biếu thức A = 49x^2 - 28x +25

\(A=49x^2-28x+25\)

\(A=\left(7x\right)^2-2.7x.2+4-4+25\)

\(A=\left(7x-2\right)^2+21\)

Vì \(\left(7x-2\right)^2\ge0\) với mọi x

\(\Rightarrow\left(7x-2\right)^2+21\ge21\) với mọi x

\(\Rightarrow Amin=21\Leftrightarrow7x-2=0\)

\(\Rightarrow7x=2\)

\(\Rightarrow x=\dfrac{2}{7}\)

Vậy \(Amin=21\Leftrightarrow x=\dfrac{2}{7}\)

\(B=8x^2-28x-1\)

\(B=2\left(4x^2-14x-\dfrac{1}{2}\right)\)

\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\left(\dfrac{7}{2}\right)^2-\dfrac{1}{2}\right]\)

\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{51}{4}\right]\)

\(B=2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\)

Vì \(2\left(2x-\dfrac{7}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\ge-\dfrac{51}{2}\)

\(\Rightarrow Bmin=-\dfrac{51}{2}\Leftrightarrow2x-\dfrac{7}{2}=0\)

\(\Rightarrow2x=\dfrac{7}{2}\)

\(\Rightarrow x=\dfrac{7}{4}\)

Vậy \(Bmin=-\dfrac{51}{2}\Leftrightarrow x=\dfrac{7}{4}\)

\(C=\left(2x^2+5\right)^2+10\)

Vì \(\left(2x^2+5\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x^2+5\right)^2+10\ge10\) với mọi x

\(\Rightarrow Cmin=10\Leftrightarrow2x^2+5=0\)

\(\Rightarrow2x^2=-5\)

\(\Rightarrow x^2=-\dfrac{5}{2}\)

\(\Rightarrow\) Không tồn tại x thỏa mãn

Vậy C không có giá trị nhỏ nhất

P/s: Câu c mình làm không có chắc nha, thấy nó sao sao ấy, không biết có sai đề không?

\(D=3x^2-8x+7\)

\(D=3\left(x^2-\dfrac{8}{3}x+\dfrac{7}{3}\right)\)

\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{16}{9}+\dfrac{7}{3}\right)\)

\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}+\dfrac{5}{9}\right)\)

\(D=3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\)

Vì \(3\left(x-\dfrac{4}{3}\right)^2\ge0\) với mọi x

\(\Rightarrow3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\ge\dfrac{5}{3}\)

\(\Rightarrow Dmin=\dfrac{5}{3}\Leftrightarrow x-\dfrac{4}{3}=0\)

\(\Rightarrow x=\dfrac{4}{3}\)

Vậy \(Dmin=\dfrac{5}{3}\Leftrightarrow x=\dfrac{4}{3}\)

\(E=x^4-2x^2+12\)

\(E=\left(x^2\right)^2-2x^2+1+11\)

\(E=\left(x^2-1\right)^2+11\)

Vì \(\left(x^2-1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x^2-1\right)^2+11\ge11\) với mọi x

\(\Rightarrow Emin=11\Leftrightarrow x^2-1=0\)

\(\Rightarrow x^2=1\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(Emin=11\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(F=4x^2+15x+2\)

\(F=\left(2x\right)^2+2.2x.\dfrac{15}{4}+\left(\dfrac{15}{4}\right)^2-\left(\dfrac{15}{4}\right)^2+2\)

\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{225}{16}+\dfrac{32}{16}\)

\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\)

Vì \(\left(2x+\dfrac{15}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\ge-\dfrac{193}{16}\)

\(\Rightarrow Fmin=-\dfrac{193}{16}\Leftrightarrow2x+\dfrac{15}{4}=0\)

\(\Rightarrow2x=-\dfrac{15}{4}\)

\(\Rightarrow x=-\dfrac{15}{4}.\dfrac{1}{2}\)

\(\Rightarrow x=-\dfrac{15}{8}\)

Vậy \(Fmin=-\dfrac{193}{16}\Leftrightarrow x=-\dfrac{15}{8}\)

\(H=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)

\(H=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)

\(H=\left(x^2+4x\right)^2-5^2\)

\(H=\left(x^2+4x\right)^2-25\)

Vì \(\left(x^2+4x\right)^2\ge0\)

\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\) với mọi x

\(\Rightarrow Hmin=-25\Leftrightarrow x^2+4x=0\)

\(\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(Hmin=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

\(I=\left(x^6+6\right)^2\)

Vì \(\left(x^6+6\right)^2\ge0\)

\(\Rightarrow Imin=0\Leftrightarrow x^6+6=0\)

\(\Rightarrow\left(x^3\right)^2=-6\)

\(\Rightarrow\) Không tồn tại x

Vậy I không có giá trị nhỏ nhất