\(x^5\) - 2\(x^4\) - (y² + 3)\(x\) + 2y² - 2 = 0

(\(x^5\) - 2\(x^4\))- (y² + 3)\(x\) + 2.(y² + 3) - 8 = 0

\(x^4\).(\(x\) - 2) - (y² + 3).(\(x\) - 2) - 8 = 0

(\(x\) - 2).(\(x^4\) - y² - 3) = 8

8 = 2³; Ư(8) = {-8; - 4; -2; - 1; 1; 2; 4; 8}

Lập bảng ta có:

vì \(x\); y nguyên nên theo bảng trên ta có các cặp \(x\); y thỏa mãn đề bài là:

(\(x\); y) = (0; -1;); (0; 1)

Lời giải:

a. $(x-2)^3+(x+2)^3-6x(x+2)(x-2)$

$=x^3-6x^2+12x-8+(x^3+6x^2+12x+8)-6x(x^2-4)$

$=2x^3+24x-6x^3+24x=-4x^3+48x$

b.

$(2x-y)^3+(2x+y)^3$

$=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3$

$=16x^3+12xy^2$

c.

$(x-2)(x+2)-(x^2+2x+4)(x-2)$

$=(x^2-4)-(x^3-2^3)=x^2-4-x^3+8=x^2-x^3+4$

Lời giải:

a. $99^3+1+3(99^2+99)=99^3+3.99^2.1+3.99.1^2+1^3=(99+1)^3=100^3=1000000$

b. $11^3-1-3(11^2-11)=11^3-3.11^2.1+3.11.1^2-1^3=(11-1)^3=10^3=1000$

a) \(70a+84b-20ab-24b^2\)

\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)

\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)

\(=\left(5a+6b\right)\left(14-4b\right)\)

\(=2\left(5a+6b\right)\left(7-2b\right)\)

b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)

\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)

c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)

\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)

\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)

\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)

\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

a, 70a + 84b - 20ab - 24b²

 = 14.(5a + 6b) - 4b(5a + 6b)

= (5a + 6b).(14 - 4b) 

Bài 4:

a. Ta thấy: $x^2-x+2=(x-\frac{1}{2})^2+1,75>0$ với mọi $x$.

Do đó để $B=\frac{x^2-x+2}{x-3}<0$ thì $x-3<0$

$\Leftrightarrow x<3$ 

b. 

$B=\frac{x(x-3)+2(x-3)+8}{x-3}=x+2+\frac{8}{x-3}$

Với $x$ nguyên, để $B$ nguyên thì $x-3$ phải là ước của 8.

$\Rightarrow x-3\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$

$\Rightarrow x\in \left\{4; 2; 5; 1; -1; 7; 11; -5\right\}$

Bài 5:

\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}=\frac{\frac{x(x+y)-y(x-y)}{(x-y)(x+y)}}{\frac{y(x+y)+x(x-y)}{(x-y)(x+y)}}\)

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