\(\widehat{zOn}-\dfrac{1}{2}\cdot\widehat{yOn}\)

\(=90^0-\widehat{zOm}-\dfrac{1}{2}\left(180^0-\widehat{xOn}\right)\)

\(=90^0-\widehat{zOm}-90^0+\dfrac{1}{2}\cdot\widehat{xOn}\)

\(=\dfrac{1}{2}\cdot\widehat{xOn}-\widehat{zOm}=\dfrac{1}{2}\left(\widehat{xOn}-\widehat{xOm}\right)=\dfrac{1}{2}\cdot\widehat{nOm}=45^0\)

Sửa đề: x-2y+3z=-33

10x=6y=5z

=>\(\dfrac{10x}{30}=\dfrac{6y}{30}=\dfrac{5z}{30}\)

=>\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}\)

mà x-2y+3z=-33

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-2y+3z}{3-2\cdot5+3\cdot6}=\dfrac{-33}{11}=-3\)

=>\(x=-3\cdot3=-9;y=-3\cdot5=-15;z=-3\cdot6=-18\)

\(\left(\dfrac{1}{6}+\dfrac{1}{3}\right)^2:\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)\)

\(=\left(\dfrac{1}{6}+\dfrac{2}{6}\right)^2:\left(\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{15}{12}\right)\)

\(=\left(\dfrac{3}{6}\right)^2:\dfrac{5}{12}=\dfrac{1}{4}\cdot\dfrac{12}{5}=\dfrac{3}{5}\)

Vì \(\widehat{A_1}\ne\widehat{B_2}\)

nên b sẽ không song song với c

mà b và c là hai đường thẳng phân biệt

nên b cắt c

\(C=\dfrac{6}{1\cdot4}+\dfrac{6}{4\cdot7}+...+\dfrac{6}{301\cdot304}\\ =2\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{301\cdot304}\right)\\ =2\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{301}-\dfrac{1}{304}\right)\\ =2\cdot\left(1-\dfrac{1}{304}\right)\\ =2\cdot\dfrac{303}{304}\\ =\dfrac{303}{152}\) 

\(B=\dfrac{11}{210}-\left(\dfrac{16}{15\cdot31}+\dfrac{13}{31\cdot44}+\dfrac{16}{44\cdot60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\left(\dfrac{1}{15}-\dfrac{1}{60}\right)\\ =\dfrac{11}{210}-\dfrac{1}{20}\\ =\dfrac{1}{420}\)

\(a.\left(\dfrac{-1}{2}\right)^2\cdot\left(\dfrac{2}{5}\right)^2 \\ =\left(\dfrac{-1}{2}\cdot\dfrac{2}{5}\right)^2\\ =\left(\dfrac{-1}{5}\right)^2\\ =\dfrac{1}{25}\\ b.\left(\dfrac{1}{9}\right)^2:\left(\dfrac{1}{3}\right)^3\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^2:\left(\dfrac{1}{3}\right)^3\\ =\left(\dfrac{1}{3}\right)^4:\left(\dfrac{1}{3}\right)^3\\ =\dfrac{1}{3}\\ c.\left(\dfrac{-1}{2}\right)^3\cdot\left(\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-1}{2}\cdot\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-3}{4}\right)^3\\ =\dfrac{-27}{64}\)

\(e.\left(\dfrac{-13}{3}-\dfrac{4}{9}\right)-\left(\dfrac{-10}{3}-\dfrac{4}{9}\right)\\ =\dfrac{-13}{3}-\dfrac{4}{9}+\dfrac{10}{3}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{3}+\dfrac{10}{3}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)\\ =-\dfrac{3}{3}=-1\\ d.\dfrac{-4}{12}-\left(-0,25-\dfrac{13}{39}\right)+0,75\\ =\dfrac{-1}{3}-\left(-\dfrac{1}{4}-\dfrac{1}{3}\right)+\dfrac{3}{4}\\ =-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{3}{4}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =0+\dfrac{4}{4}\\ =1\)

\(2,8\cdot\dfrac{-6}{13}-7,2-2,8\cdot\dfrac{7}{13}\\ =\left(2,8\cdot\dfrac{-6}{13}-2,8\cdot\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\left(\dfrac{-6}{13}-\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\dfrac{-13}{13}-7,2\\=-2,8-7,2\\ =-10\)

\(a.5\cdot3^x=5\cdot3^4\\ =>3^x=\dfrac{5\cdot3^4}{5}=3^4\\ =>x=4\\ b.7\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{7}=4^3\\ =>x=3\\ c.\dfrac{3}{5}\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{\dfrac{3}{5}}=\dfrac{35}{3}\cdot4^3\\ =>\dfrac{4^x}{4^3}=\dfrac{35}{3}\\ =>4^{x-3}=\dfrac{35}{3}\\ =>x-3=log_4\dfrac{35}{3}\\ =>x=log_4\dfrac{35}{3}+3\\ d.\dfrac{3}{2}\cdot5^x=\dfrac{3}{2}\cdot5^{12}\\ =>5^x=\dfrac{5^{12}\cdot\dfrac{3}{2}}{\dfrac{3}{2}}=5^{12}\\ =>x=12\) 

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

=>\(9\cdot5^x=9\cdot5^6\)

=>\(5^x=5^6\)

=>x=6

f: \(5\cdot3^x=7\cdot3^5-2\cdot3^5\)

=>\(5\cdot3^x=5\cdot3^5\)

=>\(3^x=3^5\)

=>x=5

g: \(5\cdot3^{x+6}=2\cdot3^5+3\cdot3^5\)

=>\(5\cdot3^{x+6}=5\cdot3^5\)

=>\(3^{x+6}=3^5\)

=>x+6=5

=>x=-1