rut gon bieu thuc
\(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(P=12\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
Rút gọn biểu thức :
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
b) P=\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
c) Q=\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
d) P = \(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(\Leftrightarrow2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow4x^2\)
Rut gon cac bieu thuc sau
b,P=\(\left(3x+1\right)^2-2\left(1+3x\right)\left(3x+5\right)+\left(3x+5\right)^2\)
T=(3+1)\(\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(P=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=\left(3x+1-3x-5\right)^2=\left(-4\right)^2=16\)
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\(T=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(\Rightarrow2T=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)
\(2T=\left(3^8-1\right)\left(3^8+1\right)=3^{16}-1\)
\(\Rightarrow T=\dfrac{3^{16}-1}{2}=21523360\)
rut gon bieu thuc sau P=\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Cho bieu thuc
\(P=\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(x+1\right)}-\frac{x^2y^2}{\left(x+1\right)\left(1-y\right)}\)
Rut gon P= x+xy-y
DKXD \(x\ne y\); \(x\ne-1\):\(y\ne1\)
Tim x y de P nguyen duong thoa man \(x^2+y^2+3xy-x-3y=0\)
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 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)
Giai hệ PT bằng phương pháp cộng
a.\(\left\{{}\begin{matrix}5.\left(x+2y\right)-3.\left(x-y\right)=99\\x-3y=7x-4y-17\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}3.\left(y-5\right)+2\left(x-3\right)=0\\7.\left(x-4\right)+3\left(x+y-1\right)=14\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2.\left(x+1\right)-5\left(y+1\right)=8\\3.\left(x+1\right)-2.\left(y+1\right)=1\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}2.\left(3y+1\right)-4\left(x-1\right)=5\\5.\left(3y+1\right)-8\left(x-1\right)=9\end{matrix}\right.\)
d: =>6y+2-4x+4=5 và 15y+5-8x+8=9
=>-4x+6y=-1 và -8x+15y=-4
=>x=-3/4; y=-2/3
c: \(\Leftrightarrow\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}3y-15+2x-6=0\\7x-28+3y+3y-3=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=21\\7x+6y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{19}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6.\left(x+y\right)=8+2x-3y\\5.\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right).\left(y-2\right)=\left(x+1\right).\left(y-3\right)\\\left(x-5\right).\left(y+4\right)=\left(x-4\right).\left(y+1\right)\end{matrix}\right.\)
rut gon bieu thuc
B= \(\frac{x^3-y^3-z^3-3xyz}{\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2}\)