12+40:(17-x)=52

=>40:(17-x)=40

=>17-x=40:40=1

=>x=17-1=16

40:( 17-x)=40

17-x=40: 40

17-x=1

x=17-1

x=16

2x-31=4646:46

=>2x-31=101

=>2x=31+101=132

=>\(x=\dfrac{132}{2}=66\)

a: \(x+40\%\cdot x=5\)

=>\(x\left(1+0,4\right)=5\)

=>1,4x=5

=>\(x=\dfrac{5}{1,4}=\dfrac{50}{14}=\dfrac{25}{7}\)

b: \(1,2x-80\%x=\dfrac{1}{4}\)

=>1,2x-0,8x=0,25

=>0,4x=0,25

=>\(x=\dfrac{0.25}{0.4}=\dfrac{25}{40}=\dfrac{5}{8}\)

d: \(x\cdot x-\dfrac{1}{9}=\dfrac{1}{3}\)

=>\(x^2=\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{4}{9}\)

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

\(372-19\cdot4+981:19-13\)

=372-76+981/19-13

\(=283+\dfrac{981}{19}=\dfrac{6358}{19}\)

Bạn làm 2 Trg họp thôi
x- 15= 0 x =15
2x-16=0 x=8

ta có 2 trương hợp

th1 x-15 =0

x=0+15

x=15

th2

2x-16=0

2x=16

x=8

\(2n-1⋮2n+3\)

=>\(2n+3-4⋮2n+3\)

=>\(-4⋮2n+3\)

=>\(2n+3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(2n\in\left\{-2;-4;-1;-5;1;-7\right\}\)

=>\(n\in\left\{-1;-2;-\dfrac{1}{2};-\dfrac{5}{2};\dfrac{1}{2};-\dfrac{7}{2}\right\}\)

\(2n+1⋮n+5\)

=>\(2n+10-9⋮n+5\)

=>\(-9⋮n+5\)

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

=>\(n\in\left\{-4;-6;-2;-8;4;-14\right\}\)

(x+2)+(x+4)+...+(x+20)=260

=>10x+(2+4+...+20)=260

=>\(10x+2\left(1+2+...+10\right)=260\)

=>\(10x+2\cdot10\cdot\dfrac{11}{2}=260\)

=>\(10x+110=260\)

=>10x=150

=>\(x=\dfrac{150}{10}=15\)

Số nhóm trong phép tỉnh tổng trên là: 

`(20 - 2) : 2 + 1 = 10` (nhóm)

`(x + 2) + (x+4) + ... + (x+20) = 260`

`=> 10x + (2+4+...+20) = 260`

`=> 10x + (20+2) . 10 = 260`

`=> 10x + 22. 10 = 260`

`=> 10x + 220 = 260`

`=> 10x = 260 - 220`

`=> 10x = 40`

`=> x = 40 : 10`

`=> x = 4`

Vậy `x = 4`

(x+3)+(x+6)+...+(x+51)=493

=>17x+(3+6+...+51)=493

=>\(17x+3\left(1+2+...+17\right)=493\)

=>\(17x+3\cdot17\cdot\dfrac{18}{2}=493\)

=>\(17x+51\cdot9=493\)

=>17x=34

=>x=2

(x+2)(y-1)=133

=>\(\left(x+2;y-1\right)\in\left\{\left(1;133\right);\left(133;1\right);\left(-1;-133\right);\left(-133;-1\right);\left(7;19\right);\left(19;7\right);\left(-7;-19\right);\left(-19;-7\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;144\right);\left(131;2\right);\left(-3;-132\right);\left(-135;0\right);\left(5;20\right);\left(17;8\right);\left(-9;-18\right);\left(-21;-6\right)\right\}\)

b: (x-5)(y-2)=21

=>\(\left(x-5;y-2\right)\in\){(1;21);(21;1);(-1;-21);(-21;-1);(3;7);(7;3);(-3;-7);(-7;-3)}

=>\(\left(x;y\right)\in\){(6;23);(26;3);(4;-19);(-16;1);(8;9);(12;5);(2;-5);(-2;-1)}

\(B=x+\dfrac{1}{2}-\left|x-\dfrac{2}{3}\right|\)
Nhận xét:

\(\left|x-\dfrac{2}{3}\right|\ge x-\dfrac{2}{3},\forall x\\ \Rightarrow-\left|x-\dfrac{2}{3}\right|\le-\left(x-\dfrac{2}{3}\right)=-x+\dfrac{2}{3},\forall x\\ \Rightarrow x+\dfrac{1}{2}-\left|x-\dfrac{2}{3}\right|\le x+\dfrac{1}{2}-x+\dfrac{2}{3}=\dfrac{7}{6},\forall x\)
hay \(B\le\dfrac{7}{6},\forall x\)

Dấu "=" xảy ra khi và chỉ khi:
\(x-\dfrac{2}{3}=0\\ \Rightarrow x=\dfrac{2}{3}\)
Vậy...