Aaaaa làm lại nha =)) tính sai .-.

Đặt 70 = x - 1 ; 34 = x - 37. Ta có :

$A=x^5-(x-1)x^4-(x-1)x^3-(x-1)x^2-(x-1)x+x-37$

$=>A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+x-37$

$=>A=2x-37=2.71-37=105$

Đặt 70 = x - 1 ; 34 = x - 37. Ta có :

$A=x^5-(x-1)x^4-(x-1)x^3-(x-1)x^2-(x-1)x+34$

$=>A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x-37$

$=>A=-37$

$x^3+y^3+z^3-3xyz$

$=(x+y)^3+z^3-3x^2y-3xy^2-3yz$

$=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)$

$=(x+y+z)(x^2+2xy+y^2-xz-yz+z^2-3xy)$

$=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$

Ta có : $(1+2x)(1-2x)-x(x+2)(x-2)$

$=1-4x^2-x(x^2-4)$

$=1-4x^2-x^3+4x$

$=-x^3-4x^2+4x+1$

$=(-x^3+x^2)-(5x^2-5x)-(x-1)$

$=(x-1).(-x^2)-5x(x-1)-(x-1)$

$=(x-1)(-x^2-5x-1)$

Ta có : $x+\dfrac{1}{2}>x-3$

* Xét $x+\dfrac{1}{2};x-3>0$

\(=>\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}x>3\\x>\dfrac{-1}{2}\end{matrix}\right.\)

$=>x>3$

* Xét $x-3;x+\dfrac{1}{2}<0$

\(=>\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\)

\(=>\left\{{}\begin{matrix}x< 3\\x< \dfrac{-1}{2}\end{matrix}\right.\)

$=>x<\dfrac{-1}{2}$

Vậy để $(x-3)(x+\dfrac{1}{2}>0$ thì \(\left[{}\begin{matrix}x>3\\x< \dfrac{-1}{2}\end{matrix}\right.\).

Viết lại đề nha bạn.

(n-1).n:2

trong đó n là số điểm đã cho

$x^9-9x^6-27$

$=[(x^3)^3-3.(x^3)^2.3+3.9.x^3-27]-27x^3$

$=(x^3)^3-27x^3$

$=(x^3-3-3x)[(x^3-3)^2+(x^3-3).3x+9x^2]$

$=(x^3-3-3x)(x^6-6x^3+9+3x^4-9x+9x^2)$

a) $(x-1)^3$=343

$=>(x-1)^3=7^3$

$=>x-1=7$

$=>x=8$

b) (số sai rồi nha bạn, ra lẻ lắm)

c) $(x-4)^2=(x-4)^4$

$=>(x-4)^4-(x-4)^2=0$

$=>(x-4)^2[(x-4)^2-1]=0$

\(=>\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}\left(x-4\right)^2=0\\\left(x-4\right)^2=1\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)

a) $x(x-y)+y(x-y)=(x+y)(x-y)=x^2-y^2$

b) $x^{n-1}(x+y)-y(x^{n-1}+y^{n-1})$

$=x^{n-1}.x+x^{n-1}y-yx^{n-1}-y.y^{n-1}$

$=x^n-y^n$