1)

a) Do \(\hept{\begin{cases}12;18;16⋮2\\A⋮2\end{cases}\Rightarrow x⋮2}\)

\(\Rightarrow x\in\left\{0;2;4;6;...\right\}\)

b) Do \(\hept{\begin{cases}12;18;16⋮2\\A⋮̸2\end{cases}}\Rightarrow x⋮̸2\)

\(\Rightarrow x\in\left\{1;3;5;7;...\right\}\)

2)

Ta có:

Do \(a:36\)dư\(16\Rightarrow a=36k+16\left(k\in N\right)\)

+ Vì \(\hept{\begin{cases}36k⋮2\\16⋮2\end{cases}\Rightarrow a⋮2}\)

+ Vì \(\hept{\begin{cases}36k⋮4\\16⋮4\end{cases}\Rightarrow a⋮4}\)

+ Vì \(\hept{\begin{cases}36k⋮18\\16⋮̸18\end{cases}\Rightarrow a⋮̸}18\)

a) 21.7² - 11.7²+90.7² + 49.125.16

= 21.7²  - 11.7² + 90 .7² + 7².125.8.2

= 21.7²-11.7²+90.7²+7².1000.2

=21.7²-11.7²+90.7²+7².2000

= 7².( 21-11 + 90 + 2000)

= 7². 2100

= 49.2100

= 102 900

b) Đặt  A = ( 2⁰+2¹+2²+2³). 2⁰.2¹.2².2³

A = (2⁰+2¹+2²+2³).2⁶

A = 2⁶+2⁷+2⁸+2⁹

=> 2A = 2⁷+2⁸+2⁹+2¹⁰

2A-A = 2¹⁰-2⁶

A = 2¹⁰-2⁶

A = 960

( 6x -39 ) .28 = 5628 

6x-39 = 5628 : 28

6x-39 = 201

6x = 201+39

6x = 162

x = 27

Vậy x = 27

(9x +2 ).3 =60

9x+2 =60:3

9x+2 = 20

9x =20-2

9x = 18

x = 2

Vậy x =2

(9x +2 ).3 =60
9x+2 =60:3
9x+2 = 20
9x =20-2
9x = 18
x = 2

\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}\)

\(\Rightarrow C=\frac{6}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\)

\(\Rightarrow C=3.\left(\frac{1}{3}-\frac{1}{47}\right)\)

\(\Rightarrow C=3.\frac{44}{141}\)

\(\Rightarrow C=\frac{44}{47}\)

\(C=\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{45.47}=3.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{45.47}\right)=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{45}-\frac{1}{47}\right)\\ \)

\(=3.\left(\frac{1}{3}-\frac{1}{47}\right)=\frac{3.44}{141}=\frac{44}{47}\)

\(A=\left\{x\inℕ|12< x< 16\right\}\)

\(\Rightarrow A=\left\{13;14;15\right\}\)

\(B=\left\{x\inℕ|x< 5\right\}\)

\(\Rightarrow B=\left\{0;1;2;3;4\right\}\)

A = { 13 ; 14; 15}

B = { 0; 1; 2; 3; 4 }

a) \(x\cdot\left(x-1\right)=0\)

\(\Rightarrow x=0\)hoặc \(x-1=0\)

                               \(x=0+1\)

                                \(x=1\)

Vậy \(x=0\)hoặc \(x=1\)

b) \(\left(x+1\right)\cdot\left(x+2\right)=0\)

\(\Rightarrow x+1=0\)hoặc \(x+2=0\)

\(\Rightarrow x=0-1\)         \(x=0-2\)

\(\Rightarrow x=-1\)             \(x=-2\)

Vậy \(x=-1\)hoặc \(x=-2\)

c) \(\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=0+1\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

\(a,x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy....

\(b,\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}}\)

Vậy....

\(c,\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy...