a) vì (x-2)² có số mũ chẵn nên (x-2)²>=0 <1>
       (2y+3)⁴có số mũ chẵn nên (2y+3)⁴=0 <2>
từ <1> và <2> suy ra :
(x-2)²=0            (2y+3)⁴=0
 x-2=0               2y+3=0
 x=2                 2y=-3
                       y=-3/2

$(x-1)^2+(y+3)^2=0$

Vì $(x-1)^2\ge0$ và $(y+3)^2\ge0$ nên:

$(x-1)^2=0$

$(y+3)^2=0$

$x-1=0$

$y+3=0$

$x=1,\ y=-3$

Vậy $(x,y)=(1,-3)$.

$2(x-5)^4+5|2y-7|^5=0$

Vì $2(x-5)^4\ge0$ và $5|2y-7|^5\ge0$ nên:

$2(x-5)^4=0$

$5|2y-7|^5=0$

$(x-5)^4=0$

$|2y-7|^5=0$

$x-5=0$

$2y-7=0$

$x=5,\ y=\dfrac{7}{2}$

Vậy $(x,y)=\left(5,\dfrac{7}{2}\right)$.

$3(x-2y)^{2004}+4\left|y+\dfrac{1}{2}\right|=0$

Vì $3(x-2y)^{2004}\ge0$ và $4\left|y+\dfrac{1}{2}\right|\ge0$ nên:

$3(x-2y)^{2004}=0$

$4\left|y+\dfrac{1}{2}\right|=0$

$x-2y=0$

$y+\dfrac{1}{2}=0$

$y=-\dfrac{1}{2}$

$x=2y=-1$

Vậy $(x,y)=\left(-1,-\dfrac{1}{2}\right)$.

$|x+3y-1|+\left(2y-\dfrac{1}{2}\right)^{2000}=0$

Vì $|x+3y-1|\ge0$ và $\left(2y-\dfrac{1}{2}\right)^{2000}\ge0$ nên:

$|x+3y-1|=0$

$\left(2y-\dfrac{1}{2}\right)^{2000}=0$

$x+3y-1=0$

$2y-\dfrac{1}{2}=0$

$y=\dfrac{1}{4}$

$x+3\cdot\dfrac{1}{4}=1$

$x=\dfrac{1}{4}$

Vậy $(x,y)=\left(\dfrac{1}{4},\dfrac{1}{4}\right)$.

Nhác quá mấy bài này hỏi làm j

Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 M=x3+x2y−2x2−xy−y2+3y+x−1M=x3+x2y−2x2−xy−y2+3y+x−1

M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)

M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1

M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1

M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1

M=x2.0+y.0+0+1M=x2.0+y.0+0+1

M=1M=1

N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2

N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)

N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2

N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2

N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2

N=x2.0−xy.0+2.0+2N=x2.0−xy.0+2.0+2

N=2N=2

P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3

P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3

P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3

P=x3.0+x2y.0−x.0+3P=x3.0+x2y.0−x.0+3

P=3

Bài 1 :

\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)

\(\Leftrightarrow x=-23\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)