a) Ư(12²) = Ư(144) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 48; 72; 144}

b) Ư(18²) = Ư(324) = {1; 2; 3; 4; 6; 9; 12; 18; 27; 36; 54; 81; 108; 162; 324}

c) Ư(24²) = Ư(576) = {1; 2; 3; 4; 6; 8; 9; 12; 16; 18; 24; 32; 36; 48; 64; 72; 96; 144; 192; 288; 576}

d) Ư(32²) = Ư(1024) = {1; 2; 4; 8; 16; 32; 64; 128; 256; 512; 1024}

;((

\(Ư\left(144\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\right\}\)

\(Ư\left(324\right)=\left\{1,2,3,4,6,9,12,18,27,36,54,81,108,162,324\right\}\)

\(Ư\left(576\right)=\left\{1,2,3,4,6,8,9,12,16,18,24,32,36,48,64,72,96,144,192,288,576\right\}\)

\(Ư\left(1024\right)=\left\{1,2,4,8,16,32,64,128,256,512,1024\right\}\)

\(Ư\left(1296\right)=\left\{1,2,3,4,6,8,9,12,16,24,27,36,48,54,72,81,108,144,162,216,324,432,648,1296\right\}\)

\(Ư\left(48\right)=\left\{1;-1;-2;2;3;-3;6;-6;8;-8;12;-12;-24;24;16;-16;48;-48\right\}\\ Ư\left(8\right)=\left\{1;-1;2;-2;-4;4;8;8\right\}\\ Ư\left(14\right)=\left\{-1;1;2;-2;7;-7;14;-14\right\}\\ Ư\left(20\right)=\left\{-1;1;2;-2;5;-5;4;-4;-10;10;-20;20\right\}\)

a,

\(\dfrac{18\left(x-y\right)^{10}}{2\left(x-y\right)^5}=9\left(x-y\right)^5\)

b, \(\dfrac{10\left(x-2\right)^{12}}{\left(2-x\right)^{10}}=\dfrac{10\left(x-2\right)^{12}}{\left(x-2\right)^{10}}=10\left(x-2\right)^2\)

c, \(\dfrac{-18\left(x-3\right)^5}{2\left(3-x\right)^3}=\dfrac{-18\left(x-3\right)^5}{-2\left(x-3\right)^3}=9\left(x-3\right)^2\)

d,\(\dfrac{x^2-6x+9}{x-3}=\dfrac{\left(x-3\right)^2}{x-3}=x-3\)

e, \(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-2x+x-2}{x+1}=\dfrac{\left(x-2\right)\left(x+1\right)}{x+1}=x-2\)

xem lại đề

a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2=4xy\)

b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\\ =6ab^2\)

a) \(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\left|x\right|+12\left|y\right|=54\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\left|y\right|=44\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}y=\pm4\\x=\pm2\end{matrix}\right.\)

Vậy ...

\(\left\{{}\begin{matrix}\left|x\right|+4\left|y\right|=18\\3\left|x\right|+\left|y\right|=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3\left|x\right|-12\left|y\right|=-54\left(1\right)\\3\left|x\right|+\left|y\right|=10\left(2\right)\end{matrix}\right.\)

Cộng (1) và (2), ta được phương trình: \(-11\left|y\right|=-44\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\)

Thay $y=4$ vào $(2)$, ta được:

\(3\left|x\right|+\left|4\right|=10\\ \Leftrightarrow3\left|x\right|=6\\ \Leftrightarrow\left|x\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Thay $y=-4$ vào $(2)$, ta được:

\(3\left|x\right|+\left|-4\right|=10\\ \Leftrightarrow3\left|x\right|+4=10\\ \Leftrightarrow3\left|x\right|=6\\ \Leftrightarrow\left|x\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy...

A = { 0 ; 1 ; 2 ; 3 ; 4 ; .... ; 50 }

B = { 1 ; 3 ; 5 ; 9 ; 15 ; 45 }

C = { 0 ; 15 ; 30 ; 45 ; 60 ; ... }

A \(\Omega\)B = { 1 ; 3 ; 5 ; 9 ; 15 ; 45 }

A \(\Omega\)C = { 0 ; 15 ; 30 ; 45 }

B \(\Omega\)C = { 15 ; 45 }

\(\hept{\begin{cases}\left(x-1\right)\left(y-2\right)-\left(x+1\right)\left(y-3\right)=4\\\left(x-3\right)\left(y+1\right)-\left(x-3\right)\left(y-5\right)=18\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy-2x-y+2-xy+3x-y+3=4\\xy+x-3y-3-xy+5x+3y-15=18\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2y=-1\\6x=36\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\x-2y=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\y=\frac{7}{2}\end{cases}}\)