\(B=2x^2+8x+1\)

\(=2\times\left(x^2+2\times x\times2+2^2-2^2+\frac{1}{2}\right)\)

\(=2\times\left[\left(x+2\right)^2-\frac{7}{2}\right]\)

\(\left(x+2\right)^2\ge0\)

\(\left(x+2\right)^2-\frac{7}{2}\ge-\frac{7}{2}\)

\(2\times\left[\left(x+2\right)^2-\frac{7}{2}\right]\ge-7\)

Vậy Min B = -7 khi x = -2

a) \(A=2\left|x-3\right|+\left|2x-10\right|=\left|2x-3\right|+\left|10-2x\right|\ge\left|2x-3+10-2x\right|=7\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(2x-3\right)\left(10-2x\right)\ge0\)\(\Leftrightarrow\)\(\frac{3}{2}\le x\le5\)

b) \(B\left|\frac{1}{4}x-8\right|+\left|2-\frac{1}{4}x\right|\ge\left|\frac{1}{4}x-8+2-\frac{1}{4}x\right|=6\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\frac{1}{4}x-8\right)\left(2-\frac{1}{4}x\right)\ge0\)\(\Leftrightarrow\)\(8\le x\le32\)

\(C=\dfrac{x^2+2-\left(x^2-2x+1\right)}{x^2+2}=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\)

\(C_{max}=1\) khi \(x=1\)

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

\(C_{min}=-\dfrac{1}{2}\) khi \(x=-2\)

Nhập Mode 7 , chạy trong khoản trung lập (-10;10)

tìm đc \(\begin{cases} C max = 1 khi x=1\\C min =-\dfrac{1}{2} khi x=-2 \end{cases}\)

Dùng cách này bạn giải trắc nghiệm sẽ nhanh hơn 

\(C\ge30\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=-1

\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)

dấu'=' xảy ra<=>x=1=>Max A=6

\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)

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

\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)

dấu"=" xảy ra<=>x=y=2=>Max B=10

\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

dấu'=' xảy ra<=>x=1,y=-3=>MinC=2

Đặt:     \(A=\left(x-3\right)\left(x+3\right)+2\left(2x+1\right)^2\)

=>       \(A=x^2-9+2\left(4x^2+4x+1\right)\)

=>       \(A=x^2-9+8x^2+8x+2\)

=>       \(A=9x^2+8x-7\)

=>       \(A=\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\)

Có:      \(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

=>      \(A\ge-\frac{79}{9}\)

DẤU "=" XẢY RA <=>     \(\left(3x+\frac{4}{3}\right)^2=0\)

<=>     \(x=-\frac{4}{9}\)

Vậy A min =     \(-\frac{79}{9}\)      <=>       \(x=-\frac{4}{9}\)

( x - 3 )( x + 3 ) + 2( 2x + 1 )²

= x² - 9 + 2( 4x² + 4x + 1 )

= x² - 9 + 8x² + 8x + 2

= 9x² + 8x - 7

= 9x² + 8x + 16/9 - 79/9

= ( 3x + 4/3 )² - 79/9

\(\left(3x+\frac{4}{3}\right)^2\ge0\forall x\Rightarrow\left(3x+\frac{4}{3}\right)^2-\frac{79}{9}\ge-\frac{79}{9}\)

Dấu " = " xảy ra <=> 3x + 4/3 = 0 => x = -4/9

=> GTNN của biểu thức = -79/9 <=> x =  -4/9