Question

помогите пожалуйста,очень нужна помощь​

Answer 1

$\displaystyle\bf\\1)\\\\a) \ x-\frac{4xy}{x+y} +y=\frac{x\cdot(x+y)-4xy+y\cdot(x+y)}{x+y} =\\\\\\=\frac{x^{2} +xy-4xy+xy+y^{2} }{x+y} =\frac{x^{2} -2xy+y^{2} }{x+y} =\frac{(x-y)^{2} }{x+y} \\\\\\b) \ x-\frac{4xy}{x-y} -y=\frac{x\cdot(x-y)-4xy-y\cdot(x-y)}{x-y} =\\\\\\=\frac{x^{2} -xy-4xy-xy+y^{2} }{x-y} =\frac{x^{2} -6xy+y^{2} }{x-y} \\\\\\c) \ \frac{(x-y)^{2} }{x+y}\cdot\frac{x^{2} -6xy+y^{2} }{x-y} =\frac{(x-y)(x^{2} -6xy+y^{2} )}{x+y}$

$\displaystyle\bf\\2)\\\\a) \ a-\frac{1-2a^{2} }{1-a} +1=\frac{a\cdot(1-a)-1+2a^{2}+1\cdot(1-a) }{1-a} =\\\\\\=\frac{a-a^{2} -1+2a^{2}+1-a }{1-a} =\frac{a^{2} }{1-a} \\\\\\b) \ 1-\frac{1}{1-a} =\frac{1-a-1}{1-a} =-\frac{a}{1-a} =\frac{a}{a-1} \\\\\\c) \ \frac{a^{2} }{1-a} :\frac{a}{a-1} =\frac{a^{2} }{1-a} \cdot\frac{a-1}{a} =-a$