Ta có : 2xy - x - y = 2

<=> 2xy - x = 2 + y

<=> x(2y - 1) = y + 2

=> x = \(\frac{y+2}{2y-1}\)

Vì x nguyên nên \(\frac{y+2}{2y-1}\) nguyên

Ta có ; \(\frac{y+2}{2y-1}=\frac{2y+4}{2y-1}=\frac{\left(2y-1\right)+5}{2y-1}=\frac{2y-1}{2y-1}+\frac{5}{2y-1}=1+\frac{5}{2y-1}\)

Để \(\frac{y+2}{2y-1}\) nguyên thì \(\frac{5}{2y-1}\) nguyên

Suy ra : 2y - 1 \(\in\) Ư(5) = {-5;-1;1;5}

Ta có bảng :

Câu 2:

TH1: x<-1,4

=>\(-x-1.1-x-1.2-x-1.3-x-1.4=5x\)

=>\(5x=-4x-5\)

=>9x=-5

=>x=-5/9(loại)

TH2: -1,4<=x<-1,3

\(\Leftrightarrow x+1.4-x-1.3-x-1.2-x-1.1=5x\)

=>5x=-2x-2,2

=>7x=-2,2

=>x=-2,2/7(loại)

TH3: -1,3<=x<-1,2

=>x+1,4+x+1,3-x-1,2-x-1,1=5x

=>5x=0,4

=>x=0,4/5(loại)

TH4: -1,2<=x<-1,1

=>-x-1,1+x+1,2+x+1,3+x+1,4=5x

=>5x=2x+2,8

=>3x=2,8

=>x=28/30(loại)

TH5: x>=-1,1

=>x+1,1+x+1,2+x+1,3+x+1,4=5x

=>5x=4x+5

=>x=5(nhận)

a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha

A=x³+1+2x+2-x³-2x=3

B=5x²+36x+7-5x²+5x=41x+7

\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)

\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)

\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)

\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)

a: =>(x+2-3)(x+2+3)=0

=>(x-1)(x+5)=0

=>x=1 hoặc x=-5

b: =>(x-1)^2=25

=>x-1=5 hoặc x-1=-5

=>x=-4 hoặc x=6

c: =>25x^2+10x+1-25x^2+9=30

=>10x+10=30

=>x+1=3

=>x=2

d: =>x^3-1-x(x^2-4)=5

=>x^3-1-x^3+4x=5

=>4x=6

=>x=3/2

\(5x\left(x-3\right)\left(x-1\right)-4x\left(x^2-2x\right)\)

\(5x^3-5x^2-15x^2+15x-4x^3+8x^2\)

\(x^3-12x^2+15x\)

\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)

\(-4x^3+16x^2-12x^2+48x-3x^3+3x^2-3x\)

\(-7x^3+7x^2+45x\)

a/ \(\left|2x-1,6\right|-2,3=1,4\)

\(\Leftrightarrow\left|2x-1,6\right|=3,7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)

Vậy ....

b/ \(5,4-\left|3x-1,2\right|=5,5\)

\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)

Mà \(\left|3x-1,2\right|\ge0\)

\(\Leftrightarrow x\in\varnothing\)

c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)

Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)

\(\Leftrightarrow x+1,3+x+2,4=4x\)

\(\Leftrightarrow2x+3,7=4x\)

\(\Leftrightarrow3,7=4x-2x\)

\(\Leftrightarrow2x=3,7\)

\(\Leftrightarrow x=1,85\)

Vậy ....

d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)

Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)

Vậy ..

a, \(\left|2x-1,6\right|-2,3=1,4\)

\(\Rightarrow\left|2x-1,6\right|=3,7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)

b,\(5,4-\left|3x-1,2\right|=5,5\)

\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)

Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)

Vậy, không có giá trị của x thỏa mãn.

c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)

\(\Leftrightarrow x+1,3+x+2,4=4x\)

\(\Leftrightarrow x+x+1,3+2,4=4x\)

\(\Leftrightarrow2x+3,7=4x\)

\(\Leftrightarrow2x-4x=-3,7\)

\(\Leftrightarrow-2x=-3,7\)

\(\Leftrightarrow x=\dfrac{3,7}{2}\)

d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)

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