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

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

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)

\(=\left(x-y-2\right)^2\)

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

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

=> x-2 \(\in\)Ư(7)= { 1,7}

Nếu x-2 = 1 => x= 1+2 => x= 3

Nếu x-2= 7 => x= 7+2 => x= 9

Nếu x= 3 thì ( x-2) ( y+1) = ( 3-2)(y+1)=7

=>  y+1 =7 => y= 7-1 => y = 6

Nếu x = 9 thì ( x- 2 )( y+1)= 7 => ( 9-2) ( y+1) =7

=> 7( y+1) =7 => y+1= 7:7 => y+1 = 1 => y= 1-1 => y=0

Vậy...

Trình bày có chỗ nào sao mong mn sửa hộ nhaaa

b) ( 2x-1)( x+3) =6

=> ( 2x-1) \(\in\)Ư( 6) = { 1: 2: 3: 6}

Mà 2x-1 là số lẻ nên 2x-1 \(\in\){ 1; 3}

Nếu 2x-1 = 1 thì 2x= 1+1 => 2x= 2 => x= 2:2 => x= 1

Nếu 2x-1 = 3 thì 2x= 1+3 => 2x=4 => x= 4:2 => x= 2

Phần còn lại làm như phần a nha :33333

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

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

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

\(=\left(x^2-1\right)\left(x-4\right)\left(x-2\right)\)

\(\left(2x-2\right).\left(3x-9\right)< 0\Leftrightarrow2\left(x-1\right).3\left(x-3\right)< 0\)

\(\Leftrightarrow6\left(x-1\right)\left(x-3\right)< 0\Leftrightarrow\orbr{\begin{cases}x-1< 0;x-3>0\\x-1>0;x-3< 0\end{cases}}\)

Mà \(x-1>x-3\Rightarrow\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 3\end{cases}}\Leftrightarrow1< x< 3\Leftrightarrow x=2\)

Vậy \(x=2\)

Vì \(\left|x^2+2x\right|\ge0;\left|y^2-9\right|\ge0\)

Dấu ''='' xảy ra <=> \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;x=-2\)

\(y^2-9=0\Leftrightarrow\left(y-3\right)\left(y+3\right)=0\Leftrightarrow y=\pm3\)

Ta có :

$\mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid =0$

Do {|x2+2x|≥0|y2−9|≥0{|x2+2x|≥0|y2−9|≥0

$\to \mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid \ge 0$

Mà $\mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid =0$

$\to$ {|x2+2x|=0|y2−9|=0{|x2+2x|=0|y2−9|=0

$\to$ $\{\begin{array}{c}{}^{}+2x=0\\ {}^{}-9=0\end{array}$

$\to$ $\{\begin{array}{c}x(x+2)=0\\ {}^{}=9\end{array}$

$\to$ $⎧⎪\; ⎪\; ⎪\; ⎪⎨⎪\; ⎪\; ⎪\; ⎪⎩\begin{array}{c}[\begin{array}{c}x=0\\ x+2=0\end{array}\\ [\begin{array}{c}y=3\\ y=-3\end{array}\end{array}$

$\to$ $⎧⎪\; ⎪\; ⎪\; ⎪⎨⎪\; ⎪\; ⎪\; ⎪⎩\begin{array}{c}[\begin{array}{c}x=0\\ x=-2\end{array}\\ [\begin{array}{c}y=3\\ y=-3\end{array}\end{array}$

Vậy $x,y\in \{0;3\};\{0;-3\};\{-2;3\};\{-2;-3\}$

đề????

TÌM x,y \(\in\) Z

1, ta có x²+y²=0

do \(x^2\ge0,y^2\ge0\Rightarrow x^2+y^2\ge\)0

mà x²+y²=0

=>x²=0,y²=0

=> x=y=0

Phần 2 tương tự nha bn 

3,|x-3|+|y+2|=0

do\(\left|x-3\right|\ge0,\left|y+2\right|\ge0\)

\(\Rightarrow\left|x-3\right|+\left|y+2\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+2\right|=0\end{cases}\Rightarrow}\hept{\begin{cases}x-3=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2\end{cases}}}\)

Phần 4 đề thiếu nhé bn

Biết điểm A(m^2-m; m^2) thuộc đồ thị hàm số y=-1/2x. Tính m

( Y+2) + ( Y+4)+........+(Y+1994)=997000

(Y + Y + .... + Y ) + (2 + 4 + ... + 1994 ) = 997000

Y x 997       +  [ (1994 + 2 ) x 997 : 2 ]    = 997000

Y x 997        +  995 006                          = 997000

Y x 997                                                   = 997000 - 995 006

Y x 997                                                    = 1994

Y                                                              = 1994 : 997

Y                                                               = 2.

Vậy Y = 2.

Hok tốt !