\(-x^3+3x^2-3x+1\) (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

Bổ sung: \(a\ne b\)

\(a^2+3a=b^2+3b\)

\(\Rightarrow a^2-b^2+3a-3b=0\)

\(\Rightarrow\left(a-b\right)\left(a+b\right)+3\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(a+b+3\right)=0\)

- Vi \(a\ne b\) nên ta chọn \(a+b+3=0\) hay \(a+b=-3\)

- Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=-3.\left[\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a^2+2ab+b^2\right)\right]\)

\(=-3.\left[\dfrac{3}{2}\left(2-3a+2-3b\right)-\dfrac{1}{2}\left(a+b\right)^2\right]\)

\(=-3.\left[\dfrac{3}{2}\left[4-3.\left(-3\right)\right]-\dfrac{1}{2}.9\right]\)

\(=-3.\left(\dfrac{3}{2}.13-\dfrac{9}{2}\right)=-3.15=-45\left(đpcm\right)\)

\(125-x^3-1\)

\(=124-x^3\)

(a+b)²=a²+2ab+b²=4 (1)

a³+b³=14 (2)

a³+b³=(a+b)(a²-ab+b²)=14

=> 2(a²-ab+b²)=14 => a²-ab+b²=7 (3)

Trừ 2 vế của (1) cho (3) => 3ab=-3=>ab= -1 

Nhân 2 vế của (1) với (2)

=> (a³+b³)(a²+2ab+b²)=14.4=56

=> a⁵+2a⁴b+a³b²+a²b³+2ab⁴+b⁵=56

=> (a⁵+b⁵)+2ab(a³+b³)+a²b²(a+b)=56

=> (a⁵+b⁵)+2.(-1).14+(-1)².2=56

=> (a⁵+b⁵)-28+2=56 => a⁵+b⁵=82

\(\left(x^2+y^2-5\right)-\left(2xy-4\right)^2=\left(x^2+y^2-1-2xy\right)^2=\left[\left(x-y\right)^2-1\right]^2=\left(x-y-1\right)^2\left(x-y+1\right)^2\)

cái này là pt đa thức thành nt

(x²+1)²-6(x²+1)²+5

=  (x²+1)²(-6+1)+5

= -5(x²+1)²+5

= -5(x⁴+2x²+1-1)

= -5(x⁴+2x²)

= -5x⁴-10x²

\(\left(x^2+1\right)^2-6\left(x^2+1\right)^2+5\\ \Leftrightarrow\left(x^2+1\right)\left(-6+5\right)\\ \Leftrightarrow\left(x^2+1\right).-1\\ \Leftrightarrow-x^2-1\)