Làm tính nhân :
a) \(\left(x+3y\right)\left(x^2-2xy+y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x\right)\)
c) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
d)\(\left(x+2\right)\left(x-1\right)\)
e) \(x\left(x-y\right)-y\left(y-x\right)\)
K
Khách
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Những câu hỏi liên quan
14 tháng 5 2017
a) \((x^2-1)(x^2+2x) =x^4+2x^3-x^2-2x\)
b) \((x+3y)(x^2-2xy+y) = x^3- 2x^2y+xy+3x^2y-6xy^2+3y^2\)
=\(x^3+xy+x^2y+6xy^2+3y^2\)
c) \((2x-1)(3x+2)(3-x)=6x^2+4x+6x-2x^2-3x-2-3+x\)= \(4x^2+8x-5\)\(\)
=
5 tháng 8 2020
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
NT
Nguyễn thành Đạt
CTVHS
8 tháng 9 2023
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
a, \(\left(x+3y\right)\left(x^2-2xy+y\right)=x^3-2x^2y+xy+3x^2y-6xy^2+3y^2\)
\(=x^3+x^2y+xy-6xy^2+3y^2\)
\(b,\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
c, \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x=17x^2-6x^3+5x-6\)
d, \(\left(x+2\right)\left(x-1\right)=x^2-x+2x-2=x^2+x-2\)
e, \(x\left(x-y\right)-y\left(y-x\right)=x^2-xy-\left(y^2-xy\right)\)
\(=x^2-xy-y^2+xy=x^2-y^2\)
Chúc bạn học tốt!!!