1, \(3x^2-5x+4\)

\(=3\left(x^2-\frac{5}{3}x\right)+1=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{23}{12}=3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\)

Ta có: \(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\ge\frac{23}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy minA = \(\frac{23}{12}\Leftrightarrow x=\frac{5}{6}\)

2, Bạn thử kiểm tra lại đề bài xem

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

Tìm giá trị nhỏ nhất của biểu thức:

\(P=3x^2+31y^2-18xy+6x-14y+2021\)

\(=3[\left(x^2-6xy+9y^2\right)+2\left(x-3y\right)+1]+\left(4y^2+4y+1\right)+2017\)

\(=3[\left(x-3y\right)^2+2\left(x-3y\right)+1]+\left(2y+1\right)^2+2017\)

\(=3\left(x-3y+1\right)^2+\left(2y+1\right)^2+2017\ge2017\)

Vậy \(MinP=2017\) khi \(\hept{\begin{cases}x-3y+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-5}{2}\\y=\frac{-1}{2}\end{cases}}\)

Thực hiện phép tính:

\(A\ge11\forall x\)

Dấu '=' xảy ra khi x=0

\(A\ge11\forall x\)

Dấu '=' xảy ra khi x=0