\(1.\left\{{}\begin{matrix}-x+3y=-10\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2y=6\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6}{-2}=-3\\x-5\cdot\left(-3\right)=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+15=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=16-15=1\end{matrix}\right.\\ 2.\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\-2x+8y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\9x=27\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\x=\dfrac{27}{9}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+y=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=7-6=1\\x=3\end{matrix}\right.\) 

\(3.\left\{{}\begin{matrix}3x-5y=-18\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5y=-18\\3x+6y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+2\cdot\left(-3\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=5+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=11\end{matrix}\right.\\ 4.\left\{{}\begin{matrix}4x+3y=-6\\2x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=-6\\4x-10y=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=-38\\2x-5y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x-5\cdot\dfrac{-38}{13}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x+\dfrac{190}{13}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x=16-\dfrac{190}{13}=\dfrac{18}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\x=\dfrac{18}{13}:2=\dfrac{9}{13}\end{matrix}\right.\)

20: \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+14y-2x-y=18-5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=13\\2x=5-y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\2x=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

21: \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\9x-3y=-24\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x+3y+9x-3y=-7-24\\3x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{31}{14}\\y=3\cdot\dfrac{-31}{14}+8=-\dfrac{93}{14}+\dfrac{112}{14}=\dfrac{19}{14}\end{matrix}\right.\)

22: \(\left\{{}\begin{matrix}-2x+y=-3\\3x+4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\3x+4\left(2x-3\right)=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-3\\3x+8x-12=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\y=2x-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=2\cdot2-3=4-3=1\end{matrix}\right.\)

23: \(\left\{{}\begin{matrix}x+y=2\\x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y-x-y=6-2\\x+y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2y=4\\x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=2-y=2-2=0\end{matrix}\right.\)

24: \(\left\{{}\begin{matrix}x-2y=-5\\3x+4y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-10\\3x+4y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y+3x+4y=-10-5\\x-2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=-15\\2y=x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2y=-3+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

25: \(\left\{{}\begin{matrix}3x-2y=12\\4x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=12\\8x+2y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-2y+8x+2y=12+10\\4x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\y=5-4x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=5-4\cdot2=5-8=-3\end{matrix}\right.\)

26: \(\left\{{}\begin{matrix}2x-y=10\\5x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y=20\\5x+2y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-2y+5x+2y=20+6\\2x-y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=26\\y=2x-10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{26}{9}\\y=2x-10=2\cdot\dfrac{26}{9}-10=\dfrac{52}{9}-10=-\dfrac{38}{9}\end{matrix}\right.\)

27: \(\left\{{}\begin{matrix}5x-2y=10\\5x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-2y-5x+2y=10-6\\5x-2y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0x=4\\2y=5x-6\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

28: \(\left\{{}\begin{matrix}3x+2y=8\\4x-3y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=-36\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x+8y-12x+9y=32+36\\3x+2y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}17y=68\\3x=8-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\3x=8-2\cdot4=8-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\)

37: \(\left\{{}\begin{matrix}2x+y=4\\2x+0y-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\y=4-2x=4-6=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

38: \(\left\{{}\begin{matrix}x-2y=2\\2x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=4\\2x-4y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x+4y=4-1\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=3\\x-2y=2\end{matrix}\right.\)

=>\(\left(x;y\right)\in\varnothing\)

39: \(\left\{{}\begin{matrix}3x+2y-2=0\\9x+6y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=2\\9x+6y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+6y=6\\9x+6y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=2\\3x+2y=2\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

40: \(\left\{{}\begin{matrix}2x-y=2\\4x-2y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=2\\4x-2y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-y=2\\2x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=0\\y=2x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=2x-2\end{matrix}\right.\)

41: \(\left\{{}\begin{matrix}x+2y=4\\2x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x+9y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+9y-2x-4y=18-8\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=10\\x=4-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2\\x=4-2\cdot2=4-4=0\end{matrix}\right.\)

42: \(\left\{{}\begin{matrix}-2x+y=-3\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+y-x-y=-3-3\\x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3x=-6\\y=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3-2=1\end{matrix}\right.\)

43: \(\left\{{}\begin{matrix}x-y=0\\2x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+2x+y=0-5\\x=y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=-5\\y=x\end{matrix}\right.\Leftrightarrow y=x=-\dfrac{5}{3}\)

44: \(\left\{{}\begin{matrix}2x+y=0\\x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x-4\cdot\left(-2x\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x=0\\y=-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-2\cdot0=0\end{matrix}\right.\)

45: \(\left\{{}\begin{matrix}-x+y=3\\x+2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y+x+2y=3+3\\x+2y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3y=6\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3-2\cdot2=3-4=-1\end{matrix}\right.\)

46: \(\left\{{}\begin{matrix}x-y=2\\3x-2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-3y=6\\3x-2y=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-3y-2x+2y=6-9\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=-3\\x=y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=3+2=5\end{matrix}\right.\)

\(a)\left(\dfrac{2}{3}\right)^5=\dfrac{2^5}{3^5}=\dfrac{32}{243}\\ \left(\dfrac{-2}{3}\right)^5=\dfrac{\left(-2\right)^5}{3^5}=\dfrac{-32}{243}\\ \left(-1\dfrac{3}{4}\right)^2=\left(-\dfrac{7}{4}\right)^2=\dfrac{\left(-7\right)^2}{4^2}=\dfrac{49}{16}\\ \left(-0,1\right)^4=\left(-\dfrac{1}{10}\right)^4=\dfrac{\left(-1\right)^4}{10^4}=\dfrac{1}{10000}\)

\(b)\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\\ \dfrac{790^4}{79^4}=\left(\dfrac{790}{79}\right)^4=10^4=10000\\ \dfrac{3^2}{15^2}=\left(\dfrac{3}{15}\right)^2=\left(\dfrac{1}{5}\right)^2=\dfrac{1^2}{5^2}=\dfrac{1}{25}\\ \dfrac{\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^{n-1}}=\left(-\dfrac{1}{2}\right)^{n-\left(n-1\right)}=\left(-\dfrac{1}{2}\right)^{n-n+1}=\left(-\dfrac{1}{2}\right)\)

Bài 15:

a: Ta có: \(\widehat{A_1}=\widehat{M_1}\)

mà hai góc này là hai góc ở vị trí so le trong

nên AB//MN

b: ta có: \(\widehat{NMC}=\widehat{MCD}\)

mà hai góc này là hai góc ở vị trí so le trong

nên MN//CD

Bài 14:

a: Ta có: \(\widehat{H_1}=\widehat{xAH}\)

mà hai góc này là hai góc ở vị trí đồng vị

nên Hm//Ax

b: Ta có: \(\widehat{A_1}=\widehat{K_1}\)

mà hai góc này là hai góc ở vị trí đồng vị

nên Ax//Kn

\(a)\left(\dfrac{6}{7}+1\dfrac{1}{2}\right)^2\\ =\left(\dfrac{6}{7}+\dfrac{3}{2}\right)^2\\ =\left(\dfrac{12}{14}+\dfrac{21}{14}\right)^2\\ =\left(\dfrac{33}{14}\right)^2\\ =\dfrac{1089}{196}\\ b)\left(2\dfrac{1}{5}-1\dfrac{2}{3}\right)^3\\ =\left(\dfrac{11}{5}-\dfrac{5}{3}\right)^3\\ =\left(\dfrac{33}{15}-\dfrac{25}{15}\right)^3\\ =\left(\dfrac{8}{15}\right)^3\\ =\dfrac{512}{3375}\\ c)3^2+4\cdot\left(\dfrac{7}{9}\right)^0+\left[\left(-5\right)^2:\dfrac{1}{5}\right]:25\\ =9+4\cdot1+\left(5^2\cdot5\right):25\\ =13+5^3:5^2\\ =13+5\\ =18\)

\(a)64^x:16^x=256\\ \Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=256\\ \Rightarrow2^{6x}:2^{4x}=256\\ \Rightarrow2^{6x-4x}=2^8\\ \Rightarrow2^{2x}=2^8\\ \Rightarrow2x=8\\ \Rightarrow x=\dfrac{8}{2}=4\\ b)\dfrac{-2401}{7^x}=-7\\ \Rightarrow7^x=\dfrac{-2401}{-7}\\ \Rightarrow7^x=343\\ \Rightarrow7^x=7^3\\ \Rightarrow x=3\\ c)\dfrac{625}{\left(-5\right)^x}=25\\ \Rightarrow\left(-5\right)^x=\dfrac{625}{25}\\ \Rightarrow\left(-5\right)^x=25\\ \Rightarrow\left(-5\right)^x=\left(-5\right)^2\\ \Rightarrow x=2\)

7 - n chia hết cho n - 2

=> (-n + 2) + 5 chia hết cho n - 2

=> -(n - 2) + 5 chia hết cho n - 2

=> 5 chia hết cho n - 2

=> n - 2 ∈ Ư(5) = {1; -1; 5; -5}

=> n ∈ {3; 1; 7; -3} 

a: Tỉ số giữa số thứ nhất và số thứ hai là:

\(\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{1}{3}\times\dfrac{2}{1}=\dfrac{2}{3}\)

b: Tỉ số giữa số thứ nhất và số thứ hai là

\(\dfrac{3}{8}:\dfrac{2}{3}=\dfrac{3}{8}\times\dfrac{3}{2}=\dfrac{9}{16}\)

a) Tỉ số của số thứ nhất và số thứ hai là: 

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

b) Tỉ số của số thứ nhất và số thứ hai là:

\(\dfrac{3}{8}:\dfrac{2}{3}=\dfrac{9}{16}\)

Lời giải:
\(B=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-99}{100}\\ =-\frac{3.8.15...99}{4.9...100}\) (do $B$ có lẻ các thừa số)

\(=-\frac{(1.3)(2.4)(3.5)...(9.11)}{2^2.3^2.4^2...10^2}\)

\(=-\frac{(1.2.3...9)(3.4.5...11)}{(2.3....10)(2.3.4...10)}\\ =-\frac{1.2.3...9}{2.3.4...10}.\frac{3.4.5...11}{2.3.4...10}\\ =-\frac{1}{10}.\frac{11}{2}=\frac{-11}{20}< \frac{-11}{21}\)

b: \(\dfrac{2}{5}-\left(\dfrac{4}{3}+\dfrac{4}{5}\right)-\left(-\dfrac{1}{9}-0,4\right)+\dfrac{11}{9}\)

\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{4}{5}+\dfrac{1}{9}+\dfrac{2}{5}+\dfrac{11}{9}\)

\(=\left(\dfrac{2}{5}-\dfrac{4}{5}+\dfrac{2}{5}\right)+\left(-\dfrac{4}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)

\(=-\dfrac{4}{3}+\dfrac{12}{9}=0\)

c: \(\dfrac{11}{8}\cdot\left[\left(-\dfrac{5}{11}:\dfrac{13}{8}-\dfrac{5}{11}:\dfrac{13}{5}\right)+\dfrac{-6}{33}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\cdot\dfrac{8}{13}-\dfrac{5}{11}\cdot\dfrac{5}{13}+\dfrac{-2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\left[-\dfrac{5}{11}\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{2}{11}\right]+\dfrac{3}{4}\)

\(=\dfrac{11}{8}\cdot\dfrac{-7}{11}+\dfrac{3}{4}=-\dfrac{7}{8}+\dfrac{3}{4}=-\dfrac{1}{8}\)

e: \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)

=>\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

=>x+1=0

=>x=-1