a, Thay x = 3 và y = -6 vào bt ta đc

\(5.3-4.\left(-6\right)=15-\left(-24\right)=39\\ b,\\ 2.\left(-2\right)^2-5.4=8-20=\left(-12\right)\\ c,\\ 5.\left(-1\right)^2+3.\left(-1\right)-1=5+\left(-3\right)-1=1\)

a) Thay x=3; y=-6

\(5x-4y=5.3-4.\left(-6\right)=15+24=39\)

b) Thay x=-2; y=4

\(2x^4-5y=2.\left(-2\right)^4-5.4=32-20=12\)

c, Thay x=0

\(5x^2+3x-1=5.0+3.0-1=-1\)

+) x=-1

\(5x^2+3x-1=5.\left(-1\right)^2+3.\left(-1\right)-1=5-3-1=1\)

+) \(x=\dfrac{1}{3}\)

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

\(=\dfrac{5}{9}+1-1=\dfrac{5}{9}\)

Giúp mình với =(((

`@` `\text {Ans}`

`\downarrow`

`a)`

`(x-1)(y+2)=7`

`=> (x - 1)(y + 2) \in` Ư`(7) = {7; 1; -1; -7}`

Ta có bảng sau:

Vậy, ta có cặp `(x; y)` thỏa mãn `{-1; 8}; {2; 5}; {-9; 0}; {-6; -3}`
`b)`

`x(y - 1) + y = 4`

`=> x(y - 1) + y - 4 = 0`

`=> x(y - 1) + (y - 1) - 3 = 0`

`=> (x + 1)(y - 1) = 3`

`=> (x + 1)(y - 1) \in` Ư`(3) = {-1; -3; 1; 3}`

Ta có bảng sau:

Vậy, ta có cặp `(x; y)` thỏa mãn `{0; 4}; {2; 2}; {-2; -2}; {-4; 0}`

a) \(\left(x+1\right)\left(y+4\right)=7\).

-Vì \(x,y\in Z\) nên ta có thể viết:

\(\left(x+1\right)\left(y+4\right)=1.7\) hay \(\left(x+1\right)\left(y+4\right)=7.1\) hay \(\left(x+1\right)\left(y+4\right)=\left(-1\right).\left(-7\right)\) hay \(\left(x+1\right)\left(y+4\right)=\left(-7\right).\left(-1\right)\)

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=1.7\):

\(\Rightarrow x+1=1\) và \(y+4=7\) 

\(\Rightarrow x=0\left(tmđk\right)\) và \(y=3\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=7.1\):

\(\Rightarrow x+1=7\) và \(y+4=1\) 

\(\Rightarrow x=6\left(tmđk\right)\) và \(y=-3\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=\left(-1\right).\left(-7\right)\):

\(\Rightarrow x+1=-1\) và \(y+4=-7\)

\(\Rightarrow x=-2\left(tmđk\right)\) và \(y=-11\left(tmđk\right)\).

+Xét trường hợp \(\left(x+1\right)\left(y+4\right)=\left(-7\right).\left(-1\right)\):

\(\Rightarrow x+1=-7\) và \(y+4=-1\)

\(\Rightarrow x=-8\left(tmđk\right)\) và \(y=-5\left(tmđk\right)\).

b) \(xy+2x-3y=-1\)

\(\Rightarrow xy+2x-3y+1=0\)

\(\Rightarrow y\left(x-3\right)=-2x-1\)

\(\Rightarrow y=-\dfrac{2x+1}{x-3}=\dfrac{2\left(x-3\right)-5}{x-3}=2-\dfrac{5}{x-3}\)

-Vì \(y\in Z\) \(\Rightarrow5⋮\left(x-3\right)\).

\(\Rightarrow\left(x-3\right)\inƯ\left(5\right)\)

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

\(\Rightarrow x\in\left\{4;2;8;-2\right\}\) (đều thỏa mãn điều kiện).

+Với \(x=4\) thì \(y=\dfrac{5}{4-3}=5\) (tmđk).

+Với \(x=2\) thì \(y=\dfrac{5}{2-3}=-5\) (tmđk).

+Với \(x=8\) thì \(y=\dfrac{5}{8-3}=1\) (tmđk)

+Với \(x=-2\) thì \(y=\dfrac{5}{-2-3}=-1\) (tmđk).

a)(x+1)(y-2)=3

x+1;y-2 thuộc Ư(3){1;-1;3;-3}

ta có bảng sau :

vậy cặp x;y thuộc {(2;3);(0;1);(4;5);(-2;-1)}
 

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

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

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

=>\(x\left(y-2\right)+y-2=1\)

=>\(\left(x+1\right)\left(y-2\right)=1\)

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

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

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)

cần kết quả đúng ko 

Tham khảo:Tìm x thuộc N , biết:a)  2x + 2x+3  =144b) (4x -1)2 =25 x 9  - Hoc24

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)