Datrys A=(a+b)^2+(a-b)^2

$$A = ( a + b ) ^ { 2 } + ( a - b ) ^ { 2 }$$

Datrys ar gyfer A

$$A=2\left(a^{2}+b^{2}\right)$$

Camau Ateb

Defnyddio'r theorem binomaidd \left(p+q\right)^{2}=p^{2}+2pq+q^{2} i ehangu'r \left(a+b\right)^{2}.

$$A=a^{2}+2ab+b^{2}+\left(a-b\right)^{2}$$

Defnyddio'r theorem binomaidd \left(p-q\right)^{2}=p^{2}-2pq+q^{2} i ehangu'r \left(a-b\right)^{2}.

$$A=a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}$$

Cyfuno a^{2} a a^{2} i gael 2a^{2}.

$$A=2a^{2}+2ab+b^{2}-2ab+b^{2}$$

Cyfuno 2ab a -2ab i gael 0.

$$A=2a^{2}+b^{2}+b^{2}$$

Cyfuno b^{2} a b^{2} i gael 2b^{2}.

$$A=2a^{2}+2b^{2}$$

Datrys ar gyfer a (complex solution)

$$a=-\frac{\sqrt{2\left(A-2b^{2}\right)}}{2}$$

$$a=\frac{\sqrt{2\left(A-2b^{2}\right)}}{2}$$

Datrys ar gyfer a

$$a=\frac{\sqrt{2\left(A-2b^{2}\right)}}{2}$$

$$a=-\frac{\sqrt{2\left(A-2b^{2}\right)}}{2}\text{, }A\geq 2b^{2}$$