\(a.\left(2-\sqrt{10}\right)\times\left(\sqrt2-\sqrt5\right)\)

\(=\sqrt2\cdot\left(\sqrt2-\sqrt5\right)\times\left(\sqrt2-\sqrt5\right)\)

\(=\sqrt2\cdot\left(\sqrt2-\sqrt5\right)^2=\sqrt2\cdot\left(2-2\sqrt{10}+5\right)\)

\(=\sqrt2\cdot\left(7-2\sqrt{10}\right)=7\sqrt2-2\sqrt{20}\)

\(=7\sqrt2-4\sqrt5\)

\(b.\sqrt3\times\left(\sqrt{72}+\sqrt{4,5}-\sqrt{12,5}\right)\)

\(=\sqrt3\times\left(6\sqrt2+\frac{3\sqrt2}{2}-\frac{5\sqrt2}{2}\right)\)

\(=6\sqrt6+\frac32\sqrt6-\frac52\sqrt6\)

\(=\sqrt6\times\left(6+\frac32-\frac52\right)\)

\(=\sqrt6\times5=5\sqrt6\)

\(c.12\times\left(\sqrt{\frac23}-\sqrt{\frac32}\right)=12\times\left(\frac{\sqrt2}{\sqrt3}-\frac{\sqrt3}{\sqrt2}\right)\)

\(=12\times\left(\frac{\sqrt6}{3}-\frac{\sqrt6}{2}\right)=12\sqrt6\times\left(\frac13-\frac12\right)\)

\(=12\sqrt6\times\left(-\frac16\right)=-2\sqrt6\)

\(\frac12x+\frac23x-1=-3\frac13\)

\(\frac76x-1=-\frac{10}{3}\)

\(\frac76x=-\frac{10}{3}+1\)

\(\frac76x=-\frac73\)

\(x=-\frac73:\frac76=-\frac73\cdot\frac67\)

\(x=-\frac63=-2\)

vậy x = -2

\(x^{2015}=x^{2016}\)

\(\Rightarrow x^{2016}-x^{2015}=0\)

\(\Rightarrow x^{2015}\left(x-1\right)=0\)

trường hợp 1:

\(x^{2015}=0\Rightarrow x=0\)

trường hợp 2:

\(x-1=0\Rightarrow x=1\)

vậy \(x\in\left\lbrace0;1\right\rbrace\)

\(x-\frac{1}{x}=5\Rightarrow\left(x-\frac{1}{x}\right)^2=5^2\)

\(x^2-2+\frac{1}{x^2}=25\)

\(x^2+\frac{1}{x^2}=25+2\)

\(x^2+\frac{1}{x^2}=27\)

vậy A = 27

A

đkxđ: x khác -3; x khác 3

\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x}{9-x^2}\)

\(\frac{\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{x^2-9}\)

\(\frac{x^3-3x^2-x^2+3x-x^3-3x^2}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

\(\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

\(-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}-\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}=0\)

\(\frac{-7x^2+3x+7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\)

\(\frac{0}{\left(x-3\right)\left(x+3\right)}=0\)

kết luận: \(x\in R\ne\pm3\)

\(\left(a+b\right)^2=4ab\)

\(a^2+2ab+b^2=4ab\)

\(a^2+2ab-4ab+b^2=0\)

\(a^2-2ab+b^2=0\)

\(\left(a-b\right)^2=0\)

⇒ a=b

\(A=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}+\frac{4}{21\cdot25}\)

\(=\frac15-\frac19+\frac19-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)

\(=\frac15-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

tổng số tuổi của 35 học sinh là:

35x12=420(tuổi)

tổng số tuổi của 35 học sinh và cô giáo là:

36x13=468(tuổi)

số tuổi của cô giáo là:

468-420=48(tuổi)

đáp số: 48 tuổi

\(\left(2a+b\right)^2-\left(b^2-2a\right)^2\)

\(=\left(2a+b-b^2+2a\right)\left(2a+b+b^2-2a\right)\)

\(=\left(-b^2+b+4a\right)\left(b^2+b\right)\)

\(=b\left(b+1\right)\left(-b^2+b+4a\right)\)