thu gọn:
\(3x^{n-2}.\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
Rút gọn các biểu thức sau :
a) \(6x^n\left(x^2-1\right)+2x^3\left(3x^{n+1}+1\right)\)
b) \(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
c) \(x^{n-3}\left(x-y\right)+y\left(x^{n-3}+x^{n-3}y^{n-1}\right)\)
\(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
\(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{2-2}\right)\)
\(=3x^{2n}-3x^{n-2}y^{n+2}+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
\(=3x^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n+2}-y^{2n}\)
\(=3x^{2n}-y^{2n}\)
P/s: Mk ko rõ đề nên làm vậy nhé!
Đề bài chắc là đơn giản tỉ lệ thức(rút gọn) nên mình làm luôn nha:
\(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
\(=3x^{2n}-3xy^{2n}+3xy^2-y^{2n}\)
\(=3x^{2n}-y^{2n}\)
\(3x^{n-2}\cdot\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\cdot\left(3x^{n-2}-y^{n-2}\right)\)=?
Thực hiện phép tính :
a, \(^{6x^n.\left(x^2-1\right)+2x.\left(3x^{n-1}+1\right)}\)
b, \(3x^{n-2}.\left(x^{n+2}y^{n+2}\right)+y^{n+2}.\left(3x^{n-2}-y^{n-2}\right)\)
c, \(4x^{n+1}-3.4^n\)
d, \(6^2.3^8.2^8-6^5.\left(6^{5-1}\right)\)
1) Tìm số tự nhiên n để đơn thức A chia hết cho đơn thức B
A= \(4x^{n+1}y^2;B=3x^3y^{n-1}\)
2) Rút gọn biểu thức
\(\left[\left(x^3+y^3\right)-2\left(x^2-y^2\right)+3\left(x+y\right)^2\right]:\left(x+y\right)\)
Câu 1:
\(\dfrac{A}{B}=\dfrac{4x^{n+1}y^2}{3x^3y^{n-1}}=\dfrac{4}{3}x^{n-2}y^{2-n+1}=\dfrac{4}{3}x^{n-2}y^{3-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}n-2>=0\\3-n>=0\end{matrix}\right.\Leftrightarrow2\le n\le3\)
Bài 2:
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\left(x-y\right)+3\left(x+y\right)^2}{x+y}\)
\(=x^2-xy+y^2-2\left(x-y\right)+3\left(x+y\right)\)
\(=x^2-xy+y^2-2x+2y+3x+3y\)
\(=x^2-xy+y^2+x+5y\)
Thực hiện phép tính :
\(3x^{n-2}.\left(x^{n+2}-y^{n+2}\right)+y^{n+2}.\left(x^{n-2}-y^{n-2}\right)\)
\(3xn^{n-2}.\left(x^{n+2}-y^{n+2}\right)+y^{n+2}.\left(3x^{n-2}-y^{n-2}\right)\)
\(=3^{2n}-3x^{n-2}y^{n+2}+3x^{n-2}y^{n-2}-y^{2n}\)
\(=3x^{2n}-y^{2n}\)
Bài 1 : Thực hiện phép tính
\(A=3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
Lời giải:
Ta có:
\(A=3x^{n-2}(x^{n+2}-y^{n+2})+y^{n+2}(3x^{n-2}-y^{n-2})\)
\(=3x^{n-2}.x^{n+2}-3x^{n-2}y^{n+2}+3y^{n+2}x^{n-2}-y^{n+2}.y^{n-2}\)
\(=3x^{n-2+n+2}-y^{n+2+n-2}=3x^{2n}-y^{2n}\)
Hãy rút gọn biểu thức:
a) \(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)+4\left(x^{n+1}+2y^{n-1}\right)\)
b) \(\left(\dfrac{3}{4}x^{n+1}-\dfrac{1}{2}y^n\right)\times2xy-\left(\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\times7xy\)
a) \(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)-4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=-8y^{n-1}+4x^{n+1}\)
b) \(\left(\dfrac{3}{4}x^{n+1}-\dfrac{1}{2}y^n\right)\cdot2xy-\left(\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}+\left(-\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}-\dfrac{14}{3}x^{n+2}y+\dfrac{35}{6}xy^{n+1}\)
\(=-\dfrac{19}{6}x^{n+2}y+\dfrac{29}{6}xy^{n+1}\)
a)\(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)+4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=4x^{n+1}-8y^{n-1}\) \(\left(=4\left(x^{n+1}-2y^{n-1}\right)\right)\)