ta có : \(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow3^2-2xy=5\)
\(\Leftrightarrow9-2xy=5\Leftrightarrow2xy=9-5=4\Leftrightarrow xy=2\)
ta có : \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=\left(3\right)^3-3.2.3=27-18=9\)
vậy \(x^3+y^3=9\) khi \(x+y=3\) và \(x^2+y^2=5\)
Đúng 0
Bình luận (0)
ta có : \(\left(x+y\right)=3\)=> \(\left(x+y\right)^2=9\)
<=> \(x^2+2xy+y^2=9\)
=> \(xy=\frac{9-\left(x^2+y^2\right)}{2}=\frac{9-5}{2}=2\)
ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=5.\left(5-2\right)=5.3=15\)
Đúng 0
Bình luận (1)
^{4}&=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4},\\[8pt](x+y)^{5}&=x^{5}+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5},\\[8pt](x+y)^{6}&=x^{6}+6x^{5}y+15x^{4}y^{2}+20x^{3}y^{3}+15x^{2}y^{4}+6xy^{5}+y^{6},\\[8pt](x+y)^{7}&=x^{7}+7x^{6}y+21x^{5}y^{2}+35x^{4}y^{3}+35x^{3}y^{4}+21x^{2}y^{5}+7xy^{6}+y^{7}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e30228a99653e6b9a65ae6640a3b7f85e2fdf144)