\(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+2\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\\ =\left(\sqrt{5}+2\right)+\left(\sqrt{3}+1\right)-\left(\sqrt{5}+\sqrt{3}\right)\\ =\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}\\ =2+1=3\)

\(-\dfrac{1}{12}-\left(-\dfrac{1}{10}\right)\\ =-\dfrac{1}{12}+\dfrac{1}{10}\\ =\dfrac{-5}{60}+\dfrac{6}{60}\\ =\dfrac{-5+6}{60}\\ =\dfrac{1}{60}\)

-1/12--1/10

=-5/60--6/60

=-11/60

$2,5\times20,21\times5\times40\times0,2$

$=(2,5\times40)\times(5\times0,2)\times20,21$

$=100\times1\times20,21$

$=100\times20,21=2021$

\(2,5\cdot20,21\cdot5\cdot40\cdot0,2\\=\left(2,5\cdot40\right)\cdot20,21\cdot\left(5\cdot0,2\right)\\ =\left(2,5\cdot4\cdot10\right)\cdot20,21\cdot1\\ =100\cdot20,21\\ =2021\)

\(\sqrt{\dfrac{9}{4}}-\sqrt{2}+\sqrt{2}\\ =\dfrac{3}{2}-\left(\sqrt{2}-\sqrt{2}\right)\\ =\dfrac{3}{2}-0\\ =\dfrac{3}{2}\)

\(\left(7-\dfrac{1}{5}+\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\\ =7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\\ =\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)\\ =1-2-1\\ =-2\)

\(\left(7-\dfrac{1}{5} +\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\)

\(=7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\)

\(=\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)-\dfrac{1}{3}\)

\(=1-2+\left(-1\right)-\dfrac{1}{3}\)

\(=\left[1+\left(-1\right)\right]-2-\dfrac{1}{3}\)

\(=0-2-\dfrac{1}{3}\)

\(=-2-\dfrac{1}{3}\)

\(=-\dfrac{6}{3}-\dfrac{1}{3}\)

\(=-\dfrac{7}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.\left(2^2.5\right)}\)

\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

\(H=\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\sqrt{3}\\ =\dfrac{4\left(1+\sqrt{3}\right)}{-2}-\sqrt{3}\\ =-2\left(1+\sqrt{3}\right)-\sqrt{3}\\ =-2-2\sqrt{3}-\sqrt{3}\\ =-2-3\sqrt{3}\)

Cảm ơn bạn nhé

\(1-\left(4\dfrac{2}{5}+x-7\dfrac{2}{3}\right):15\dfrac{1}{3}=0\\ 1-\left(\dfrac{22}{5}+x-\dfrac{23}{3}\right):\dfrac{46}{3}=0\\ 1-\left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=0\\ \left(\dfrac{-49}{15}+x\right):\dfrac{46}{3}=1\\ -\dfrac{49}{15}+x=\dfrac{46}{3}\\ x=\dfrac{46}{3}+\dfrac{49}{15}\\ x=\dfrac{279}{15}=\dfrac{93}{5}\)

$1-\left(4.\frac25+x-\frac{7.2}{3}\right):15.\frac13=0$

$\Rightarrow \left(\frac85-\frac{14}{3}+x\right):15:3=1$

$\Rightarrow \left(-\frac{46}{15}+x\right):15=3$

$\Rightarrow -\frac{46}{15}+x=3.15$

$\Rightarrow -\frac{46}{15}+x=45$

$\Rightarrow x=45-\left(-\frac{46}{15}\right)=\frac{721}{15}$

Trung bình cộng của 6 số chẵn bằng 47. Suy ra 2 số giữa là: 46 và 48.
Vậy, 6 số cần tìm: 42,44,46,48,50,52

Xét tứ giác ABCD có \(\widehat{ABC}+\widehat{ADC}=180^0\)

nên ABCD là tứ giác nội tiếp

=>\(\widehat{DAC}=\widehat{DBC};\widehat{BAC}=\widehat{BDC}\)

mà \(\widehat{CDB}=\widehat{CBD}\)(CB=CD)

nên \(\widehat{DAC}=\widehat{BAC}\)

=>AC là phân giác của góc BAD