\(K=5x^2+4xy+y\left(y-4\right)-10x\)

\(K=5x^2+4xy+y^2-4y-10x\)

\(K=\left(4x^2+4xy+y^2\right)+x^2-4y-10x\)

\(K=\left[\left(2x+y\right)^2-2\left(2x+y\right).2+4\right]+\left(x^2-2x+1\right)-5\)

\(K=\left(2x+y-2\right)^2+\left(x-1\right)^2-5\)

Mà  \(\left(2x+y-2\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow K\ge-5\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}2x+y-2=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(K_{Min}=-5\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

\(K=5x^2+4xy+y\left(y-4\right)-10x.\)

\(=\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2-2x+1\right)-1\)

\(=\left(\left(2x\right)^2+y^2+2^2+2.2x.y-2.2x.2-2.y.2\right)+\left(x^2-2x+1\right)-1\)

\(=\left(2x+y-2\right)^2+\left(x-1\right)^2-1\ge-1\)

Dấu "=" xảy ra khi

\(\hept{\begin{cases}\left(2x+y-2\right)^2=0\\\left(x-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}}\)

Câu trên T làm sai rồi. Quên để ý phía trước có cộng thêm 4. MinK=-5

\(H=x^2+\left(x-2\right)\left(3x-1\right)\)

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

\(=4x^2-7x+2\)

\(=\left(2x\right)^2-2\cdot2\cdot\frac{7}{4}x+\left(\frac{7}{4}\right)^2-\frac{17}{16}\)

\(=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\)

Vì \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge-\frac{17}{16}\forall x\)

Dấu " = " xảy ra khi và chỉ khi \(\left(2x-\frac{7}{4}\right)^2=0\)

\(\Leftrightarrow x=\frac{7}{8}\)

Vậy \(H_{min}=-\frac{17}{16}\)tại \(x=\frac{7}{8}\)

\(x^2+\left(x-2\right)\left(3x-1\right)=x^2+3x^2-x-6x+2=4x^2-7x+2\)

\(=4x^2-7x+\frac{49}{16}-\frac{17}{16}\)

\(=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\)

Vì: \(\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge\frac{17}{16}\forall x\)

=> Min H =17/16 tại \(\left(2x-\frac{7}{4}\right)^2=0\Rightarrow x=\frac{7}{8}\)

=.= hok tốt!!

\(H=x^2+\left(x-2\right)\left(3x-1\right)=x^2+3x^2-x-6x+2=4x^2-7x+2\)

                               \(=4x^2-2.2x.\frac{7}{4}+\frac{49}{16}-\frac{17}{16}=\left(2x-\frac{7}{4}\right)^2-\frac{17}{16}\ge\frac{-17}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow2x-\frac{7}{4}=0\Leftrightarrow x=\frac{7}{8}\)

Vậy H_Min = -17/16 khi và chỉ khi x = 7/8

GTLN:lớn lắm

GTNN:3

\(C=x^2-4x+8\)

\(C=x^2-4x+4+4\)

\(C=\left(x-4\right)^2+4\ge4\)

Dấu bằng xảy ra 

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Vậy Min A = 4 <=> x= 4

giải lun câu này dùm ik: D= x- x^2+ 3. Tìm GTNN hoặc GTLN

\(A=\dfrac{8x+3}{4x^2+1}=\dfrac{4\left(4x^2+1\right)-\left(4x-1\right)^2}{4x^2+1}=4-\dfrac{\left(4x-1\right)^2}{4x^2+1}\le4\)

Vậy GTLN của A là 4 . Dấu " = " xảy ra khi \(\left(4x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{4}\)

\(\text{a)* }A=\dfrac{8x+3}{4x^2+1}=\dfrac{\left(4x^2+8x+4\right)-\left(4x^2+1\right)}{4x^2+1}\\ =\dfrac{4x^2+8x+4}{4x^2+1}-\dfrac{4x^2+1}{4x^2+1}=\dfrac{4\left(x+1\right)^2}{4x^2+1}-1\ge-1\)

Dấu \("="\) xảy ra khi \(\left(x+1\right)^2=0\)

\(\Leftrightarrow x=-1\)

\(\text{* }A=\dfrac{8x+3}{4x^2+1}=\dfrac{-\left(16x^2-8x+1\right)+\left(16x^2+4\right)}{4x^2+1}\\ =\dfrac{-\left(16x^2-8x+1\right)}{4x^2+1}+\dfrac{16x^2+4}{4x^2+1}\\ =\dfrac{-\left(16x^2-8x+1\right)}{4x^2+1}+\dfrac{4\left(4x^2+1\right)}{4x^2+1}\\ =\dfrac{-\left(4x-1\right)^2}{4x^2+1}+4\)

Dấu \("="\) xảy ra khi \(4x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy \(A_{Min}=-1\Leftrightarrow x=-1\)

\(A_{Max}=4\Leftrightarrow x=\dfrac{1}{4}\)

\(B=\dfrac{x}{\left(x+2010\right)^2}\overset{AM-GM}{\ge}\dfrac{x}{8020x}=\dfrac{1}{8020}\)

Dấu "=" xảy ra khi: \(x=2010\)