Thực hiện phép tính
\(\left(x^2y-\frac{1}{2}xy+y^2\right).\left(x-\frac{1}{2}y\right)-x^2y\left(x-\frac{1}{2}\right)\)
Thực hiện phép tính sau
a)\(\left(x^2y^2-\frac{1}{2}xy+2y\right)\left(x-2y\right)\)
b)\(\left(x^2-xy+y^2\right)\left(x+y\right)\)
a) biết chết liền
b) \(\left(x^2-xy+y^2\right)\left(x+y\right)=x^3+y^3\)
thực hiện phép tính:
a/ \(\left(x^2y^2-\frac{1}{2}xy+2y\right)\left(x-2y\right)\)
b/ \(\left(y-1\right)\left(y^2+y+1\right)+\left(\frac{1}{3}x^3y-y\right)\left(2x+y^2\right)\)
a/ (\(x^3y^2\)-\(\frac{1}{2}x^3y\) + \(2xy\) - \(2x^2y^3\) + \(xy^2\) - \(4y^2\) =
Thực hiện phép tính
\(\frac{3xy^2+x^2y}{xy\left(x-y\right)}\)- \(\frac{3x^2y+xy^2}{xy\left(x-y\right)}\)
\(ĐKXĐ:x\ne y,x\ne0,y\ne0\)
Ta có : \(\frac{3xy^2+x^2y}{xy\left(x-y\right)}-\frac{3x^2y+xy^2}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y-3x^2y-xy^2}{xy.\left(x-y\right)}\)
\(=\frac{-3xy.\left(x-y\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}=\frac{-2xy.\left(x-y\right)}{xy.\left(x-y\right)}=-2\)
\(\frac{3xy^2+x^2y}{xy\left(x-y\right)}-\frac{3x^2y+xy^2}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y}{xy\left(x-y\right)}+\frac{-\left(3x^2y+xy^2\right)}{xy.\left(x-y\right)}\)
\(=\frac{3xy^2+x^2y-3x^2y-xy^2}{xy.\left(x-y\right)}\)
\(=\frac{\left(3xy^2-3x^2y\right)+\left(x^2y-xy^2\right)}{xy.\left(x-y\right)}\)
\(=\frac{3xy.\left(y-x\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}\)
\(=\frac{-3xy.\left(x-y\right)+xy.\left(x-y\right)}{xy.\left(x-y\right)}\)
\(=\frac{\left(x-y\right).\left(-3xy+xy\right)}{xy.\left(x-y\right)}\)
\(=\frac{-3xy+xy}{xy}\)
\(=\frac{-2xy}{xy}\)
\(=-2.\)
\(\frac{3xy^2+x^2y}{xy\left(x-y\right)}-\frac{3x^2y+xy^2}{xy\left(x-y\right)}\)( ĐKXĐ : \(x\ne y;x,y\ne0\))
= \(\frac{3xy^2+x^2y-3x^2y-xy^2}{xy\left(x-y\right)}\)
\(=\frac{xy^2\left(3-1\right)+x^2y\left(1-3\right)}{xy\left(x-y\right)}\)
\(=\frac{xy^2\left(3-1\right)-x^2y\left(3-1\right)}{xy\left(x-y\right)}\)
\(=\frac{2\left(xy^2-x^2y\right)}{xy\left(x-y\right)}\)
\(=\frac{2xy\left(y-x\right)}{-xy\left(y-x\right)}\)
\(=-2\)
a)thực hiện phép tính: A = \(\left(\frac{x}{y^2-xy}+\frac{y}{x^2-xy}\right):\left(\frac{x^2-y^2}{x^2y+xy^2}\right)\)
b)rút gọn biểu thức B = \(\frac{\left|x\right|+\left|y\right|}{x+y}\)
thực hiện phép tính:
\(\left(x^2y^2-\frac{1}{2}xy+2y\right).\left(x-2y\right)\)
Bài 1: Thu gọn
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)
d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)
e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)
f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)
g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)
h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)
k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)
n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)
m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)
p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
giải hệ phương trình
1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)
3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)
8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(E=\frac{2}{x\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
mann nào trả lời đc thui k hết 5 cái nick lun :D
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
Bài 1 rút gọn biểu thức
A=\(\left(x-\frac{4xy}{x+y}+y\right)\):\(\left(\frac{x}{x+y}-\frac{y}{x-y}-\frac{2xy}{x^2-y^2}\right)\)
B=\(\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right)\):\(\left(\frac{x^2+4x^2y^2+y^4}{x^2+y+xy+x}\right):\left(\frac{1}{2x^2+y+2}\right)\)