Resolver (x-1/5y)^2-(8/15y+11/2x)^2+(9/2x+frac{2}{3}y)^2-[(frac{1}{5}y-3x)*(3x+frac{1}{5}y)+(-frac{2}{5}y)^2]

$$( x - \frac { 1 } { 5 } y ) ^ { 2 } - ( \frac { 8 } { 15 } y + \frac { 11 } { 2 } x ) ^ { 2 } + ( \frac { 9 } { 2 } x + \frac { 2 } { 3 } y ) ^ { 2 } - [ ( \frac { 1 } { 5 } y - 3 x ) \cdot ( 3 x + \frac { 1 } { 5 } y ) + ( - \frac { 2 } { 5 } y ) ^ { 2 } ]$$

Avaliar

$$-\frac{4xy}{15}$$

Passos Para a Resolução

Utilize o teorema binomial \left(a-b\right)^{2}=a^{2}-2ab+b^{2} para expandir \left(x-\frac{1}{5}y\right)^{2}.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Utilize o teorema binomial \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para expandir \left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}\right)+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Para calcular o oposto de \frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}, calcule o oposto de cada termo.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\frac{64}{225}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine \frac{1}{25}y^{2} e -\frac{64}{225}y^{2} para obter -\frac{11}{45}y^{2}.

$$x^{2}-\frac{2}{5}xy-\frac{11}{45}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{2}{5}xy e -\frac{88}{15}yx para obter -\frac{94}{15}xy.

$$x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine x^{2} e -\frac{121}{4}x^{2} para obter -\frac{117}{4}x^{2}.

$$-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Utilize o teorema binomial \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para expandir \left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}.

$$-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\frac{81}{4}x^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{117}{4}x^{2} e \frac{81}{4}x^{2} para obter -9x^{2}.

$$-9x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{94}{15}xy e 6xy para obter -\frac{4}{15}xy.

$$-9x^{2}-\frac{4}{15}xy-\frac{11}{45}y^{2}+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{11}{45}y^{2} e \frac{4}{9}y^{2} para obter \frac{1}{5}y^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Considere \left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right). A multiplicação pode ser transformada na diferença dos quadrados através da regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y\right)^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(\frac{1}{5}y\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}\right)^{2}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Calcule \frac{1}{5} elevado a 2 e obtenha \frac{1}{25}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(3x\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-3^{2}x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Calcule 3 elevado a 2 e obtenha 9.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(-\frac{2}{5}y\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}\right)^{2}y^{2}\right)$$

Calcule -\frac{2}{5} elevado a 2 e obtenha \frac{4}{25}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\frac{4}{25}y^{2}\right)$$

Combine \frac{1}{25}y^{2} e \frac{4}{25}y^{2} para obter \frac{1}{5}y^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{5}y^{2}-9x^{2}\right)$$

Para calcular o oposto de \frac{1}{5}y^{2}-9x^{2}, calcule o oposto de cada termo.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\frac{1}{5}y^{2}+9x^{2}$$

Combine \frac{1}{5}y^{2} e -\frac{1}{5}y^{2} para obter 0.

$$-9x^{2}-\frac{4}{15}xy+9x^{2}$$

Combine -9x^{2} e 9x^{2} para obter 0.

$$-\frac{4}{15}xy$$

Expandir

$$-\frac{4xy}{15}$$

Passos Para a Resolução

Utilize o teorema binomial \left(a-b\right)^{2}=a^{2}-2ab+b^{2} para expandir \left(x-\frac{1}{5}y\right)^{2}.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Utilize o teorema binomial \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para expandir \left(\frac{8}{15}y+\frac{11}{2}x\right)^{2}.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\left(\frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}\right)+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Para calcular o oposto de \frac{64}{225}y^{2}+\frac{88}{15}yx+\frac{121}{4}x^{2}, calcule o oposto de cada termo.

$$x^{2}-\frac{2}{5}xy+\frac{1}{25}y^{2}-\frac{64}{225}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine \frac{1}{25}y^{2} e -\frac{64}{225}y^{2} para obter -\frac{11}{45}y^{2}.

$$x^{2}-\frac{2}{5}xy-\frac{11}{45}y^{2}-\frac{88}{15}yx-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{2}{5}xy e -\frac{88}{15}yx para obter -\frac{94}{15}xy.

$$x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}-\frac{121}{4}x^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine x^{2} e -\frac{121}{4}x^{2} para obter -\frac{117}{4}x^{2}.

$$-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Utilize o teorema binomial \left(a+b\right)^{2}=a^{2}+2ab+b^{2} para expandir \left(\frac{9}{2}x+\frac{2}{3}y\right)^{2}.

$$-\frac{117}{4}x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+\frac{81}{4}x^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{117}{4}x^{2} e \frac{81}{4}x^{2} para obter -9x^{2}.

$$-9x^{2}-\frac{94}{15}xy-\frac{11}{45}y^{2}+6xy+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{94}{15}xy e 6xy para obter -\frac{4}{15}xy.

$$-9x^{2}-\frac{4}{15}xy-\frac{11}{45}y^{2}+\frac{4}{9}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Combine -\frac{11}{45}y^{2} e \frac{4}{9}y^{2} para obter \frac{1}{5}y^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right)+\left(-\frac{2}{5}y\right)^{2}\right)$$

Considere \left(\frac{1}{5}y-3x\right)\left(3x+\frac{1}{5}y\right). A multiplicação pode ser transformada na diferença dos quadrados através da regra: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}y\right)^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(\frac{1}{5}y\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\left(\frac{1}{5}\right)^{2}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Calcule \frac{1}{5} elevado a 2 e obtenha \frac{1}{25}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-\left(3x\right)^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(3x\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-3^{2}x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Calcule 3 elevado a 2 e obtenha 9.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}y\right)^{2}\right)$$

Expanda \left(-\frac{2}{5}y\right)^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\left(-\frac{2}{5}\right)^{2}y^{2}\right)$$

Calcule -\frac{2}{5} elevado a 2 e obtenha \frac{4}{25}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{25}y^{2}-9x^{2}+\frac{4}{25}y^{2}\right)$$

Combine \frac{1}{25}y^{2} e \frac{4}{25}y^{2} para obter \frac{1}{5}y^{2}.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\left(\frac{1}{5}y^{2}-9x^{2}\right)$$

Para calcular o oposto de \frac{1}{5}y^{2}-9x^{2}, calcule o oposto de cada termo.

$$-9x^{2}-\frac{4}{15}xy+\frac{1}{5}y^{2}-\frac{1}{5}y^{2}+9x^{2}$$

Combine \frac{1}{5}y^{2} e -\frac{1}{5}y^{2} para obter 0.

$$-9x^{2}-\frac{4}{15}xy+9x^{2}$$

Combine -9x^{2} e 9x^{2} para obter 0.

$$-\frac{4}{15}xy$$