\(7x^2-x+1=7\left(x^2-\dfrac{x}{7}+\dfrac{1}{196}\right)+\dfrac{27}{28}\)

\(=7\left(x-\dfrac{1}{14}\right)^2+\dfrac{27}{28}\ge\dfrac{27}{28}\forall x\)

\(Min=\dfrac{27}{28}\Leftrightarrow x=\dfrac{1}{14}\)

A=   7x² - x + 1 

A=  7( x² - 2.1/14x + 1/196) + 27/28 

A= 7(x - 1/14)² + 27/28

A = 7(x - 1/14)² ≥ 0  ⇔ 7(x-1/14)² +27/28 ≥ 27/28 

A(min)= 27/28 ⇔ x = 1/14

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

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

Ta có :

4x³+6x²+4x+1

= (4x³+2x²)+(4x²+2x)+(2x+1)

= 2x²(2x+1)+2x(2x+1)+(2x+1)

= (2x²+2x+1)(2x+1)

a) \(A=\left(37^3+12^3\right):49-37\times12\)

\(=\left(37+12\right)\left(37^2+12^2-37\times12\right):49-37\times12\)

\(=37^2+12^2-2\times37\times12\)

\(=\left(37-12\right)^2=25^2=625\)

b) \(B=\left(52^3-48^3\right):4+52\times48\)

\(=\left(52-48\right)\left(52^2+48^2+52\times48\right):4+52\times48\)

\(=52^2+48^2+2\times52\times48\)

\(=\left(52+48\right)^2=100^2=10000\)

\(-x^3+3x^2-3x+1\) (tại x=6, ta có)

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

\(=6\left(-6^2+3.6-3\right)+1\)

\(=6\left(-36+18-3\right)+1\)

\(=6.\left(-21\right)+1=-125\)

Ta có :

-x³+3x²-3x+1

= -(x³-1)+(3x²-3x)

= -(x-1)(x²+x+1)+3x(x-1)

=(x-1)(3x-x²-x-1)

= (x-1)(-x²+2x-1)

= (x-1)[-(x²-x)+(x-1)]

= (x-1)[-x(x-1)+(x-1)]

= (x-1)².(1-x) (1)

Thay x = 6 vào (1) ta được :

-x³+3x²-3x+1=(x-1)².(1-x)

= (6-1)².(1-6)

= 5².(-5) = -125

Ta có:

\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-zx=b\\z^2-xy=c\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\)

⇒\(ax+by+cz=x^3+y^3+z^3-3xyz\)

                        \(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)

                        \(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)

                        \(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(x+y\right)z-3xy\right]\)

⇒\(ax+by+cz⋮x+y+z\)

em ghi sai đề hay sao á

\(a,=5xy\left(3x-y+4x^2\right)\)

\(b,=\left(x-4\right)^2-16=\left(x-4+4\right)\left(x-4-4\right)\)

\(=x\left(x-8\right)\)

\(c,=\left(x-5\right)^2-9^2=\left(x-14\right)\left(x+4\right)\)

\(d,=\left(x-y\right)^2-\left(x+5\right)^2=\left(2x-y+5\right)\left(-y-5\right)\)

a) \(15x^2y-5xy^2+20x^3y=5xy\left(3x-y+4x^2\right)\)

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

c) \(x^2+25-10x-81=\left(x-5\right)^2-9^2=\left(x-14\right)\left(x+4\right)\)

d) \(x^2-2xy+y^2-x^2-10x-25=\left(x-y\right)^2-\left(x+5\right)^2=-\left(2x-y+5\right)\left(y+5\right)\)

\(Với\) \(\left\{{}\begin{matrix}x=8\\y=9\end{matrix}\right.\) \(thay\) \(vào\), \(ta\) \(có\):

\(A=x\left(x-y\right)+y\left(x+y\right)\)

\(=8\left(8-9\right)+9\left(8+9\right)\)

\(=8.-1+9.17=-8+153=145\)

x(x-y)+y(x+y)

=x²-xy+xy+y²

=x²+y²

Thay x=8; y=9 vào biểu thức trên ta được:

x²+y²

=8²+9²

=64+81

=145