\(\Leftrightarrow4xy-4x-11y+121=121\)

\(\Leftrightarrow4x\left(y-1\right)-11\left(y-1\right)=121\)

\(\Leftrightarrow\left(4x-11\right)\left(y-1\right)=121\)

Pt ước số, chắc em tự giải nốt được

Bài 1:

\(a,=11\left(x+y\right)+x\left(x+y\right)=\left(x+11\right)\left(x+y\right)\\ b,=225-\left(2x+y\right)^2=\left(15-2x-y\right)\left(15+2x+y\right)\)

Bài 2:

\(A=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ A=\left(72-2\right)\left(120-2\right)=70\cdot118=8260\)

Bài 3:

\(a,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\\ \Leftrightarrow24x+25=49\\ \Leftrightarrow24x=24\Leftrightarrow x=1\)

thk you very much UwU

4x+11y = 4xy

<=> 4x - 4xy - 11 + 11y = -11

<=> 4x( 1 - y) - 11(1 - y) = -11

<=> (4x-11)(y - 1) = 11

<=> 4x - 11 và y - 1 là Ư(11) = {-11;-1;1;11}

Bảng giá trị

Vậy pt có tập nghiệp (x,y) là (0;0); (2,5;-10); (3; 12); (5,5; 2)

B3) a) x(x-5)-4(x-5)=0

<=> (x-4)(x-5)=0

TH1 :x-4=0

<=.x=4

TH2 : x-5=0

<=>x=5

b) x(x-6)-7x-42=0

<=>x(x+6)-7(x+6)=0

<=>(x-7)(x+6)=0

th1;x-7=0

<=>x=7

th2; x+6=0

<=>x=-6

c)x^3-5x^2+x-5=0

<=>  x(x^2+1)-5(x^2+1)=0

<=> (x-5)(x^2+1)=0

th1:x-5=0

<=>x=5

TH2 : x^2+1=0

<=> x^2=-1 ( vo li )

=> th2 ko tồn tại 

nho thick nha  

Bài 3

a, x(x-5)-4(x-5)=0

 (x-4)(x-5)=0

=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

b,x(x+6)-7(x+6)=0

(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)

c,x^2(x-5)+(x-5)=0

(x^2+1)(x-5)=0

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(y=\frac{1}{x^2+\sqrt{x}}\)

\(y=\frac{1}{x^2+\sqrt{x}}\)

Ta có \(4x+11y=4xy\Leftrightarrow4xy-4x-11y+11=11\Leftrightarrow4x\left(y-1\right)-11\left(y-1\right)=11\Leftrightarrow\left(y-1\right)\left(4x-11\right)=11\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y-1=1\\4x-11=11\end{matrix}\right.\\\left\{{}\begin{matrix}y-1=11\\4x-11=1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2\\x=\dfrac{11}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}y=12\\x=3\end{matrix}\right.\end{matrix}\right.\)

Vì x,y nguyên dương

Vậy (x;y)=(3;12)

\(4xy-3x+2y=15\)

\(\Leftrightarrow x\left(4y-3\right)+2y=15\)

\(\Leftrightarrow2x\left(4y-3\right)+4y=30\)

\(\Leftrightarrow2x\left(4y-3\right)+4y-3=27\)

\(\Leftrightarrow\left(2x+1\right)\left(4x-3\right)=27\)

Ta có bảng sau: 

Vậy: \(\left(x;y\right)=\left\{\left(13;1\right);\left(-1;-6\right);\left(1;3\right);\left(-5;0\right)\right\}\)

4xy - 3x + 2y = 15

(4xy + 2y) - 3x = 15

2y(2x + 1) - 3x = 15

4y(2x + 1) - 6x = 30

4y(2x + 1) - 6x - 3 = 30 - 3

4y(2x + 1) - 3(2x + 1) = 27

(2x + 1)(4y - 3) = 27

*) TH1: 2x + 1 = -27; 4y - 3 = -1

+) 2x + 1 = -27

2x = -28

x = -14

+) 4y - 3 = -1

4y = 2

y = 1/2 (loại)

*) TH2: 2x + 1 = -9; 4y - 3 = -3

+) 2x + 1 = -9

2x = -10

x = -5

+) 4y - 3 = -3

4y = 0

y = 0

*) TH3: 2x + 1 = -1; 4y - 3 = -27

+) 2x + 1 = -1

2x = -2

x = -1

+) 4y - 3 = -27

4y = -24

y = -6

*) TH4: 2x + 1 = -3; 4y - 3 = -9

+) 2x + 1 = -3

2x = -4

x = -2

+) 4y - 3 = -9

4y = -6

y = -3/2 (loại)

*) TH5: 2x + 1 = 1; 4y - 3 = 27

+) 2x + 1 = 1

2x = 0

x = 0

+) 4y - 3 = 27

4y = 30

y = 15/2 (loại)

*) TH6: 2x + 1 = 3; 4y - 3 = 9

+) 2x + 1 = 3

2x = 2

x = 1

+) 4y - 3 = 9

4y = 12

y = 3

*) TH7: 2x + 1 = 9; 4y - 3 = 3

+) 2x + 1 = 9

2x = 8

x = 4

+) 4y - 3 = 3

4y = 6

y = 3/2 (loại)

*) TH8: 2x + 1 = 27; 4y - 3 = 1

+) 2x + 1 = 27

2x = 26

x = 13

+) 4y - 3 = 1

4y = 4

y = 1

Vậy ta tìm được các cặp giá trị (x; y) thỏa mãn:

(13; 1); (1; 3); (-1; -6); (-5; 0)