\(B=\dfrac{5^2\cdot6^{11}\cdot\left(-16\right)^2+6^2\cdot\left(-12\right)^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)

\(=\dfrac{5^2\cdot3^{11}\cdot2^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2}{2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot\left(2^6\cdot3\cdot5\right)^3}\)

\(=\dfrac{5^2\cdot3^{11}\cdot2^{19}+5^2\cdot3^{10}\cdot2^{14}}{2^{17}\cdot3^{12}\cdot5^4-3^{11}\cdot2^{18}\cdot5^3}\)

\(=\dfrac{5^2\cdot3^{10}\cdot2^{14}\left(3\cdot2^5+1\right)}{2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)}\)

\(=\dfrac{1}{5}\cdot\dfrac{1}{3}\cdot\dfrac{1}{8}\cdot\dfrac{97}{13}=\dfrac{97}{1560}\)

a: \(2x=\dfrac{1}{3}\)

=>\(x=\dfrac{1}{3}:2=\dfrac{1}{3}\cdot\dfrac{1}{2}=\dfrac{1}{6}\)

b: \(2x+\dfrac{1}{2}=-\dfrac{5}{3}\)

=>\(2x=-\dfrac{5}{3}-\dfrac{1}{2}=\dfrac{-13}{6}\)

=>\(x=-\dfrac{13}{6}:2=-\dfrac{13}{12}\)

c: \(-3x-\dfrac{3}{4}=\dfrac{5}{6}\)

=>\(-3x=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{10+9}{12}=\dfrac{19}{12}\)

=>\(x=-\dfrac{19}{12}:3=-\dfrac{19}{36}\)

d: \(\dfrac{3}{4}x+\dfrac{1}{2}=\dfrac{-3}{5}\)

=>\(\dfrac{3}{4}x=-\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{-11}{10}\)

=>\(x=-\dfrac{11}{10}:\dfrac{3}{4}=\dfrac{-11}{10}\cdot\dfrac{4}{3}=\dfrac{-44}{30}=\dfrac{-22}{15}\)

e: \(-\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{4}\)

=>\(\dfrac{1}{2}x=-\dfrac{5}{3}-\dfrac{3}{4}=\dfrac{-20-9}{12}=\dfrac{-29}{12}\)

=>\(x=-\dfrac{29}{12}\cdot2=-\dfrac{29}{6}\)

\(a.2x=\dfrac{1}{3}\\ x=\dfrac{1}{3}:2\\ x=\dfrac{1}{6}\\ b.2x+\dfrac{1}{2}=-\dfrac{5}{3}\\ 2x=\dfrac{-5}{3}-\dfrac{1}{2}\\ 2x=-\dfrac{13}{6}\\ x=-\dfrac{13}{6}:2\\ x=-\dfrac{13}{12}\\ c.-3x-\dfrac{3}{4}=\dfrac{5}{6}\\ -3x=\dfrac{5}{6}+\dfrac{3}{4}\\ -3x=\dfrac{19}{12}\\ x=\dfrac{19}{12}:\left(-3\right)\\ x=-\dfrac{19}{36}\)

\(d.\dfrac{3}{4}x+\dfrac{1}{2}=\dfrac{-3}{5}\\ \dfrac{3}{4}x=\dfrac{-3}{5}-\dfrac{1}{2}\\ \dfrac{3}{4}x=-\dfrac{11}{10}\\ x=-\dfrac{11}{10}:\dfrac{3}{4}\\ x=-\dfrac{22}{15}\\ e.\dfrac{-5}{3}-\dfrac{1}{2}x=\dfrac{3}{4}\\ \dfrac{1}{2}x=-\dfrac{5}{3}-\dfrac{3}{4}\\ \dfrac{1}{2}x=-\dfrac{29}{12}\\ x=\dfrac{-29}{12}:\dfrac{1}{2}\\ x=-\dfrac{29}{6}\)

\(A=\dfrac{3^{2022}+2}{3^{2022}-1}=\dfrac{3^{2022}-1+3}{3^{2022}-1}=1+\dfrac{3}{3^{2022}-1}\)

\(B=\dfrac{3^{2022}}{3^{2022}-3}=\dfrac{3^{2022}-3+3}{3^{2022}-3}=1+\dfrac{3}{3^{2022}-3}\)

Vì \(3^{2022}-1>3^{2022}-3\)

nên \(\dfrac{3}{3^{2022}-1}< \dfrac{3}{3^{2022}-3}\)

=>\(1+\dfrac{3}{3^{2022}-1}< 1+\dfrac{3}{3^{2022}-3}\)

=>A<B

\(x\cdot\dfrac{5}{36}-\dfrac{5}{6}=\dfrac{-3}{4}\)

=>\(x\cdot\dfrac{5}{36}=\dfrac{-3}{4}+\dfrac{5}{6}=\dfrac{-9+10}{12}=\dfrac{1}{12}\)

=>\(x=\dfrac{1}{12}:\dfrac{5}{36}=\dfrac{1}{12}\cdot\dfrac{36}{5}=\dfrac{3}{5}\)

2xy-5x+2y-14=0

=>2xy+2y-5x-5-9=0

=>2y(x+1)-5(x+1)=9

=>(x+1)(2y-5)=9

=>\(\left(x+1\right)\left(2y-5\right)=1\cdot9=\left(-1\right)\cdot\left(-9\right)=\left(-9\right)\cdot\left(-1\right)=9\cdot1=3\cdot3=\left(-3\right)\cdot\left(-3\right)\)

=>\(\left(x+1;2y-5\right)\in\left\{\left(1;9\right);\left(-1;-9\right);\left(-9;-1\right);\left(9;1\right);\left(3;3\right);\left(-3;-3\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;7\right);\left(-2;-2\right);\left(-10;2\right);\left(8;3\right);\left(2;4\right);\left(-4;1\right)\right\}\)

a: M thuộc tia AB nên M nằm giữa A và B hoặc B nằm giữa A và M

b: Các tia đối nhau gốc N là tia NA và tia NC

Các tia trùng nhau là AN và AC

Lời giải:

Gọi $d=ƯCLN(n-5,n-2)$

$\Rightarrow n-5\vdots d; n-2\vdots d$

$\Rightarrow (n-2)-(n-5)\vdots d$

$\Rightarrow 3\vdots d$

Để ps tối giản thì $d\neq 3$

Điều này xảy ra khi $n-2\not\vdots 3$

$\Leftrightarrow n\neq 3k+2$ với mọi $k$ tự nhiên, $k\neq 0$

G nằm giữa hai điểm E và F

nên GE và GF là hai tia đối nhau

\(2\left[3-9\cdot\left(-3\right)+2\left(5-7\right)\right]-18:\left(-3\right)^2\)

\(=2\left[3+27+2\cdot\left(-2\right)\right]-18:9\)

\(=2\left[30-4\right]-2\)

\(=2\cdot26-2=50\)

=2.(3-9.-3 +2.-2)-18:(-3)² 

=2.(3--27+-4)-18:-9

=2.(30+-4) -18:-9

=2.26-18:-9

=52--2

=54

\(\left(\dfrac{2}{3}x-27\right)\cdot\dfrac{3}{2}=-90\)

=>\(\dfrac{2}{3}x-27=-60\)

=>\(\dfrac{2}{3}x=-33\)

=>\(x=-33:\dfrac{2}{3}=-\dfrac{99}{2}\)