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x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a-b\right)^{2}=a^{2}-2ab+b^{2} per desenvolupar \left(x-\frac{1}{5}y\right)^{2}.
x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}\right)+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}.
x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\frac{64}{225}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Per trobar l'oposat de \frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}, cerqueu l'oposat de cada terme.
x^{2}-\frac{2}{5}xy-\frac{11}{45}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu \frac{1}{25}y^{2} i -\frac{64}{225}y^{2} per obtenir -\frac{11}{45}y^{2}.
x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{2}{5}xy i -\frac{88}{15}yx per obtenir -\frac{94}{15}xy.
-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu x^{2} i -\frac{121}{4}x^{2} per obtenir -\frac{117}{4}x^{2}.
-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\frac{81}{4}x^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}.
-9x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{117}{4}x^{2} i \frac{81}{4}x^{2} per obtenir -9x^{2}.
-9x^{2}-\frac{4}{15}xy-\frac{11}{45}y^{2}+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{94}{15}xy i 6xy per obtenir -\frac{4}{15}xy.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{11}{45}y^{2} i \frac{4}{9}y^{2} per obtenir \frac{1}{5}y^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y\right)^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Considereu \left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right). La multiplicació es pot transformar en una diferència de quadrats fent servir la regla: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}\right)^{2}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Expandiu \left(\frac{1}{5}y\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Calculeu \frac{1}{5} elevat a 2 per obtenir \frac{1}{25}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-3^{2}x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Expandiu \left(3x\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Calculeu 3 elevat a 2 per obtenir 9.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}\right)^{2}y^{2}\right)
Expandiu \left(-\frac{2}{5}y\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\frac{4}{25}y^{2}\right)
Calculeu -\frac{2}{5} elevat a 2 per obtenir \frac{4}{25}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{5}y^{2}-9x^{2}\right)
Combineu \frac{1}{25}y^{2} i \frac{4}{25}y^{2} per obtenir \frac{1}{5}y^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\frac{1}{5}y^{2}+9x^{2}
Per trobar l'oposat de \frac{1}{5}y^{2}-9x^{2}, cerqueu l'oposat de cada terme.
-9x^{2}-\frac{4}{15}xy+9x^{2}
Combineu \frac{1}{5}y^{2} i -\frac{1}{5}y^{2} per obtenir 0.
-\frac{4}{15}xy
Combineu -9x^{2} i 9x^{2} per obtenir 0.
x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a-b\right)^{2}=a^{2}-2ab+b^{2} per desenvolupar \left(x-\frac{1}{5}y\right)^{2}.
x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}\right)+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}.
x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\frac{64}{225}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Per trobar l'oposat de \frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}, cerqueu l'oposat de cada terme.
x^{2}-\frac{2}{5}xy-\frac{11}{45}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu \frac{1}{25}y^{2} i -\frac{64}{225}y^{2} per obtenir -\frac{11}{45}y^{2}.
x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{2}{5}xy i -\frac{88}{15}yx per obtenir -\frac{94}{15}xy.
-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu x^{2} i -\frac{121}{4}x^{2} per obtenir -\frac{117}{4}x^{2}.
-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\frac{81}{4}x^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Utilitzeu el teorema del binomi \left(a+b\right)^{2}=a^{2}+2ab+b^{2} per desenvolupar \left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}.
-9x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{117}{4}x^{2} i \frac{81}{4}x^{2} per obtenir -9x^{2}.
-9x^{2}-\frac{4}{15}xy-\frac{11}{45}y^{2}+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{94}{15}xy i 6xy per obtenir -\frac{4}{15}xy.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)
Combineu -\frac{11}{45}y^{2} i \frac{4}{9}y^{2} per obtenir \frac{1}{5}y^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y\right)^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Considereu \left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right). La multiplicació es pot transformar en una diferència de quadrats fent servir la regla: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}\right)^{2}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Expandiu \left(\frac{1}{5}y\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Calculeu \frac{1}{5} elevat a 2 per obtenir \frac{1}{25}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-3^{2}x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Expandiu \left(3x\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)
Calculeu 3 elevat a 2 per obtenir 9.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}\right)^{2}y^{2}\right)
Expandiu \left(-\frac{2}{5}y\right)^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\frac{4}{25}y^{2}\right)
Calculeu -\frac{2}{5} elevat a 2 per obtenir \frac{4}{25}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{5}y^{2}-9x^{2}\right)
Combineu \frac{1}{25}y^{2} i \frac{4}{25}y^{2} per obtenir \frac{1}{5}y^{2}.
-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\frac{1}{5}y^{2}+9x^{2}
Per trobar l'oposat de \frac{1}{5}y^{2}-9x^{2}, cerqueu l'oposat de cada terme.
-9x^{2}-\frac{4}{15}xy+9x^{2}
Combineu \frac{1}{5}y^{2} i -\frac{1}{5}y^{2} per obtenir 0.
-\frac{4}{15}xy
Combineu -9x^{2} i 9x^{2} per obtenir 0.