b.
\(=x^2+2xy+y^2-2xy-2xy=\left(x+y\right)^2-4xy=7^2-4.6=49-24=25\)
a, \(\dfrac{2y+2x}{xy}=\dfrac{2\left(x+y\right)}{xy}=\dfrac{14}{6}=\dfrac{7}{3}\)
b, \(\left(x+y\right)^2-4xy=49-24=25\)
c, \(\left(x+y\right)^3-3xy\left(x+y\right)==343-3.6.7=217\)
d, \(\left\{{}\begin{matrix}y=\dfrac{6}{x}\\x+\dfrac{6}{x}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6}{x}\\x^2+6-7x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1;y=6\\\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\end{matrix}\right.\)
Với x = 1 ; y = 6 thì
\(\dfrac{1+36}{\left|1-6\right|}=\dfrac{37}{5}\)
Với x = 6 ; y = 1 thì \(\dfrac{36+1}{\left|6-1\right|}=\dfrac{37}{5}\)
mik sẽ biến đổi r bn tự thay số vào nha
a) dễ r bn tự làm nha
b) \(x^2+y^2-2xy\)
\(=\left(x+y\right)^2-2xy-2xy=\left(x+y\right)^2-4xy\)
c) \(x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left(x^2+y^2-xy\right)\)
d) \(\dfrac{x^2+y^2}{\left|x-y\right|}\)
\(=\dfrac{\left(x+y\right)^2-2xy}{\sqrt{x-y}^2}\)
\(=\dfrac{\left(x+y\right)^2-2xy}{\sqrt{x^2-2xy+y^2}}\)
\(=\dfrac{\left(x+y\right)^2-2xy}{\sqrt{x^2+y^2-2xy}}\)
`a)`
\(\dfrac{2}{x}+\dfrac{2}{y}=\dfrac{2x+2y}{xy}=\dfrac{2\left(x+y\right)}{xy}=\dfrac{2.7}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)
`b)`\(x^2+y^2-2xy=\left(x+y\right)^2-4xy=7^2-4.6=49-24=25\)
`c)`\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(x+y\right)^3-3xy\left(x+y\right)=7^3-3.6.7=343-126=217\)
