12x³+4x²-27x-9=(12x³+4x²)-(27x-9)=4x²(3x+1)-3²(3x+1)=(3x+1)(4x²-3²)

cau b mjk chua ra 

bn thiếu rồi 

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

Bài 4:

a) Ta có: \(x^3+6x^2+12x+8\)

\(=x^3+2x^2+4x^2+8x+4x+8\)

\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+4x+4\right)\)

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

b) Ta có: \(x^3-3x^2+3x-1\)

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

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

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

\(=\left(x-1\right)^3\)

c) Ta có: \(1-9x+27x^2-27x^3\)

\(=1-3x-6x+18x^2+9x^2-27x^3\)

\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)

\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)

\(=\left(1-3x\right)^3\)

d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)

\(=\left(x+\frac{1}{2}\right)^3\)

e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)

\(=\left(3x-2y\right)^3\)

a)= \(\frac{\left(2x+3\right)^2}{2x^2+3x-4x-6}\)

=\(\frac{\left(2x+3\right)^2}{x\left(2x+3\right)-2\left(2x+3\right)}\)

= \(\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(2x+3\right)}\)

=\(\frac{2x+3}{x-2}\)

b) = \(\frac{3\left|x-4\right|}{3x^2-3x-1296}\)

= \(\frac{3\left|x-4\right|}{3\left(x^2-x-432\right)}\)

=\(\frac{\left|x-4\right|}{x^2-x-432}\)

ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

                          \(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)

                      \(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............

a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)

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

b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)

c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)

d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)

a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)

b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)

c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)

d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)

chiều rộng của hình chữ nhật là:

56,14 x 3/4 = 42,105 (m)

chu vi của hình chữ nhật là:

(56,14 + 42,105) x 2 = 196,39 (m)

đ\s.......