\(A=x^4+6x^3+13x^2+12x+12\)

     \(=\left(x^4+6x^3+19x^2+30x+25\right)-6x^2-18x-30+17\)

      \(=\left(x^4+6x^3+19x^2+30x+25\right)-6\left(x^2+3x+5\right)+17\)

       \(=\left(x^2+3x+5\right)^2-6\left(x^2+3x+5\right)+17\)

Đặt \(t=x^2+3x+5\)

Khi đó \(A=t^2-6t+17=t^2-2.t.3+9+8=\left(t-3\right)^2+8\ge8\)

Dấu "=" xảy ra <=> t - 3 = 0 <=> t = 3

                                          <=> \(x^2+3x+5=3\Leftrightarrow x^2+3x+2=0\)

                                           \(\Leftrightarrow x^2+x+2x+2=0\)

                                            \(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)

                                             \(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy A_Min = 8 khi và chỉ khi x = -1 hoặc x = -2

CÁC BẠN GIẢI NHANH HỘ NHÚN VỚI

\(A=\left(x^4-3x^3+2x^2\right)-3\left(x^3-3x^2+2x\right)+2\left(x^2-3x+2\right)+2019\)

\(=x^2\left(x^2-3x+2\right)-3x\left(x^2-3x+2\right)+2\left(x^2-3x+2\right)+2019\)

\(=\left(x^2-3x+2\right)\left(x^2-3x+2\right)+2019\)

\(=\left(x^2-3x+2\right)^2+2019\ge2019\)

\(A_{min}=2019\) khi \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a)\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\5x-13=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{13}{5}\end{array}\right.\)

b)\(6x^4=9x^3\Leftrightarrow6x^4-9x^3=0\Leftrightarrow3x^3\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x^3=0\\2x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)

c)\(\left(x-2\right)^2-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x-2\right)^2=4x^2+12x+9\)

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

\(\Leftrightarrow x-2=2x+3\)

\(\Leftrightarrow-x=5\Leftrightarrow x=-5\)

1/

a/ \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)

\(D=2x\left[10\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)\right]-5x\left[4\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\right]\)

\(D=20x\left(x^2-\frac{1}{2}x-\frac{1}{5}\right)-20x\left(x^2-\frac{1}{2}x-\frac{1}{4}\right)\)

\(D=20x^3-10x^2-4x-20x^3+10x^2+5x\)

\(D=x\)

b/ Mình xin sửa lại đề:

Tính giá trị biểu thức \(E\left(x\right)=x^5-13x^4+13x^3-13x^2+13x+2012\)

Tại x = 12

\(E\left(x\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x-1\right)x+2012\)

\(E\left(x\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+2012\)

\(E\left(x\right)=2012-x\)

\(E\left(x\right)=2000\)

2/

a/ \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

<=> \(2x^2-10x-3x-2x^2=26\)

<=> \(-13x=26\)

<=> \(x=-2\)

b/ Bạn vui lòng coi lại đề.

3a/ Ta có \(D=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)

\(D=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)

\(D=-10\)

Vậy giá trị của D không phụ thuộc vào x (đpcm)

Giúp mik vs^^

Thank 

\(Sửa:A=x^4-6x^3+13x^2-12x+2021\\ A=\left(x^4-6x^3+9x^2\right)+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x\right)^2+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x+2\right)^2+2017\ge2017\\ A_{min}=2017\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

a: \(x^2-8x+21=x^2-8x+16+5=\left(x-4\right)^2+5>=5\)

Dấu '=' xảy ra khi x=4

b: \(16x^2+16x-30\)

\(=16x^2+2\cdot4x\cdot2+4-34\)

\(=\left(4x+2\right)^2-34>=-34\)

Dấu '=' xảy ra khi x=-1/2

d: \(-x^2+12x+34\)

\(=-\left(x^2-12x-34\right)\)

\(=-\left(x^2-12x+36-70\right)\)

\(=-\left(x-6\right)^2+70< =70\)

Dấu '=' xảy ra khi x=6