Question

помогите пожалуйстаааа​

Answer 1

$\displaystyle\bf\\41)\\\\1) \ \ \frac{4-4x^{2} }{2x+2} :\frac{4x^{2} -4x}{3} =\frac{4-4x^{2} }{2x+2}\cdot\frac{3}{4x^{2} -4x} =\frac{4\cdot(1-x^{2} )}{2\cdot(x+1)} \cdot\frac{3}{4x\cdot(x-1)} =\\\\\\=\frac{4\cdot(1-x)\cdot(1+x)\cdot 3}{2\cdot(x+1)\cdot4x\cdot(x-1)} =-\frac{3}{2x} \\\\\\2) \ \ \frac{(4x-3)^{2}+24x }{8x^{2} +6x} -\frac{3}{2x} =\frac{16x^{2} -24x+9+24x}{2x(4x+3)} -\frac{3}{2x} =$

$\displaystyle\bf\\=\frac{16x^{2} +9}{2x(4x+3)}-\frac{3}{2x}=\frac{16x^{2} +9-3\cdot(4x+3)}{2x(4x+3)} =\frac{16x^{2} +9-12x-9}{2x(4x+3)} =\\\\\\=\frac{16x^{2} -12x}{2x(4x+3)} =\frac{4x\cdot(4x-3)}{2x(4x+3)} =\frac{8x-6}{4x+3}$

$\displaystyle\bf\\42)\\\\1) \ \ \frac{2x+3}{x^{2} -6x+9} :\frac{4x^{2} -9}{9-x^{2} } =\frac{2x+3}{(x-3)^{2} } \cdot\frac{9-x^{2} }{4x^{2} -9}=\\\\\\=\frac{2x+3}{(x-3)^{2} } \cdot\frac{(3-x)(3+x)}{(2x-3)(2x+3)} =\frac{3+x}{(3-x)(2x-3)} \\\\\\2) \ \ \frac{3+x}{(3-x)(2x-3)} +\frac{1}{2x-3}=\frac{3+x+3-x}{(3-x)(2x-3)} =\frac{6}{(3-x)(2x-3)} \\\\\\3) \ \ \frac{6}{(3-x)(2x-3)} :(2x-6)=\frac{6}{(3-x)(2x-3)} \cdot\frac{1}{2\cdot(x-3)} =\\\\\\=-\frac{3}{(x-3)^{2}(2x-3) }$