Question

Можете решить качественно вот эти задание по математике 8 класс

Answer 1

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Answer 2

$\displaystyle\bf\\2)\\\\\sqrt{1+\sqrt{2+\sqrt{x} } } =2\\\\\Big(\sqrt{1+\sqrt{2+\sqrt{x} } } \Big)^{2} =2^{2} \\\\1+\sqrt{2+\sqrt{x} } =4\\\\\sqrt{2+\sqrt{x} } =3\\\\\Big(\sqrt{2+\sqrt{x} } \Big)^{2} =3^{2} \\\\2+\sqrt{x} =9\\\\\sqrt{x} =7\\\\\boxed{x=49}$

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

$\displaystyle\bf\\4)\\\\a) \ \frac{3!+4!+5!}{6!-5!} =\frac{3!\cdot(1+4+4\cdot 5)}{5!\cdot(6-1)} =\frac{3!\cdot25}{5!\cdot5} =\frac{3!\cdot 5}{3!\cdot 4\cdot 5} =0,25\\\\\\b) \ C_{5}^{3}+C_{10} ^{8} =\frac{5!}{3!\cdot(5-3)!} +\frac{10!}{8!\cdot(10-8)!} =\\\\\\=\frac{3!\cdot 4\cdot 5}{3!\cdot 2!} +\frac{8!\cdot 9\cdot 10}{8!\cdot 2!} =\frac{4\cdot 5}{1\cdot 2} +\frac{9\cdot 10}{1\cdot 2} =10+45=55$