played

is riding

will be

won't go

Does..........do

Câu 1:

\(M=132-\left\{100-\left[\left(78-73\right)^2:5+9\right]\right\}\)

\(=132-\left\{100-\left[5^2:5+9\right]\right\}\)

\(=132-100+5+9\)

=32+14

=46

Câu 2:

105-[(2x+7)-13]=25

=>(2x+7)-13=105-25=90

=>2x-6=90

=>2x=96

=>x=96/2=48

Câu 4:

\(N=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

Câu 5:
\(xy-x+2y=3\)
\(\Rightarrow xy-x+2y-2=3-2\)
\(\Rightarrow\left(xy-x\right)+\left(2y-2\right)=1\)
\(\Rightarrow x\left(y-1\right)+2\left(y-1\right)=1\)
\(\Rightarrow\left(x+2\right)\left(y-1\right)=1\)
Ta có bảng sau:

Vậy các cặp (x;y) là (-1;2) ; (-3;0)
#YM

Bài 3:

Để A là số nguyên thì \(2n+7⋮n+3\)

=>\(2n+6+1⋮n+3\)

=>\(1⋮n+3\)

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

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

Bài 4:

Để A là số nguyên thì \(2n+7⋮n+1\)

=>\(2n+2+5⋮n+1\)

=>\(5⋮n+1\)

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

=>\(n\in\left\{0;-2;4;-6\right\}\)

Bài 5:

Để A là số nguyên thì \(6n-3⋮3n+1\)

=>\(6n+2-5⋮3n+1\)

=>\(-5⋮3n+1\)

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

=>\(3n\in\left\{0;-2;4;-6\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

mà n nguyên 

nên \(n\in\left\{0;-2\right\}\)

Bài 6:
Để A là số nguyên thì \(3n+4⋮n-1\)

=>\(3n-3+7⋮n-1\)

=>\(7⋮n-1\)

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

=>\(n\in\left\{2;0;8;-6\right\}\)