Giups mik vs

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

\(=x^2-2x+1+1\)

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a/ \(4x^2+4x+11\)

\(=\left(2x^2\right)+2\cdot2x+1-1+11\)

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

\(=\left(2x+1\right)^2+10\)

Có :  \(\left(2x+1\right)^2\ge0\)

\(\Rightarrow\left(2x+1\right)^2+10\ge10\)

\(\Rightarrow GTNN\left(4x^2+4x+11\right)=10\)

   Với \(\left(2x+1\right)^2=0;x=-\frac{1}{2}\)

\(a,A=4x^2+4x+11\)

\(A=(2x+1)^2+10\)

Do \((2x+1)^2\ge0\Rightarrow(2x+1)^2+10\ge10\forall x\)

\(\Rightarrow Min_a=10\Rightarrow2x+1=0\Rightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

Vậy giá trị nhỏ nhất của A là 10 khi x = -1/2

\(c,\)\(C=x^2-2x+y^2-4y+7\)

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

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

\(\Rightarrow C_{min}=2\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x-1=0\\y-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\y=2\end{cases}}}\)

a) \(A=1-8x-x^2=-\left(x^2+8x+16\right)+17=-\left(x-4\right)^2+17\le17\)

\(ĐTXR\Leftrightarrow x=4\)

b) \(B=5-2x+x^2=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

\(ĐTXR\Leftrightarrow x=1\)

c) \(C=x^2+4y^2-6x+8y-2021=\left(x^2-6y+9\right)+\left(4y^2+8y+4\right)-2034=\left(x-3\right)^2+\left(2y+2\right)^2-2034\ge-2034\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

a: Ta có: \(A=-x^2-8x+1\)

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

\(=-\left(x^2+8x+16-17\right)\)

\(=-\left(x+4\right)^2+17\le17\forall x\)

Dấu '=' xảy ra khi x=-4

b: Ta có: \(x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

a) Ta có: \(A=4x^2+4x+11\)

        \(\Rightarrow A=4x^2+2x+2x+11\)

        \(\Rightarrow A=2x.\left(2x+1\right)+\left(2x+1\right)+10\)

        \(\Rightarrow A=\left(2x+1\right).\left(2x+1\right)+10\)

        \(\Rightarrow A=\left(2x+1\right)^2+10\)

  Ta lại có: \(\left(2x+1\right)^2\ge0\forall x\inℝ\)

             \(\Rightarrow A\ge10\)

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

                        \(\Rightarrow2x+1=0\)

                        \(\Rightarrow2x=-1\)

                        \(\Rightarrow x=\frac{-1}{2}\)

      Vậy \(A_{min}=10\Leftrightarrow x=\frac{-1}{2}\)

( x - 1) ( x + 6 ) ( x + 2 ) ( x + 3 ) 

<=> ( x² + 6x - x - 6 ) ( x² + 3x + 2x + 6)

<=> ( x² - 5x )² lun nhỏ hơn 0 

Nên dấu " =" xảy ra khi ( x²- 5x)² = 0

x² - 5x= 0 <=> x ( x - 5) = 0 <=> x=0 hoặc 5 

^^ Học tốt nha!!!!

a) Ta có : 4x² + 4x + 11

= (2x)² + 4x + 11

= (2x)² + 4x + 1 + 10

= (2x + 1)² + 10

Mà (2x + 1)² \(\ge0\forall x\)

Nên (2x + 1)² + 10 \(\ge10\forall x\)

Vậy GTNN của biểu thức là : 10 khi x = \(-\frac{1}{2}\)