Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

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

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

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

\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

a,ta có : x^3+y^3-xy(x+y)=(x+y)(x^2+y^2-xy) -xy(x+y)=(x+y)(x^2+y^2-2xy=(x+y)(x-y)^2

(đpcm)

b)x^3-y^3+xy(x-y)=(x-y)(x^2+y^2+xy)+xy(x-y)=(x-y)(x^2+y^2+2xy)=(x-y)(x+y)^2 (đpcm)

h) Sửa lại đề bài chút xíu:

$(xy+ab)^2+(ay-bx)^2=x^2y^2+a^2b^2+2abxy+a^2y^2-2aybx+b^2x^2$

$=x^2y^2+a^2b^2+a^2y^2+b^2x^2$

$=(x^2y^2+b^2x^2)+(a^2b^2+a^2y^2)$

$=x^2(y^2+b^2)+a^2(b^2+y^2)=(a^2+x^2)(b^2+y^2)$

j)

$ab(x^2+y^2)+xy(a^2+b^2)=abx^2+aby^2+xya^2+xyb^2$
$=(abx^2+xya^2)+(aby^2+xyb^2)$

$=ax(bx+ay)+by(ay+bx)=(ax+by)(ay+bx)$

k)

$(xy-ab)^2+(bx+ay)^2=x^2y^2-2xyab+a^2b^2+b^2x^2+2bxay+a^2y^2$

$=x^2y^2+a^2b^2+b^2x^2+a^2y^2$

$=(x^2y^2+b^2x^2)+(a^2b^2+a^2y^2)=x^2(y^2+b^2)+a^2(b^2+y^2)$

$=(a^2+x^2)(b^2+y^2)$

e)

$x^2-(2a+b)xy+2aby^2=x^2-2axy-bxy+2aby^2$

$=x(x-2ay)-by(x-2ay)=(x-by)(x-2ay)$

g)

$y^2-(3a+2b)xy+6abx^2=(y^2-2bxy)-(3axy-6abx^2)$

$=y(y-2bx)-3ax(y-2bx)=(y-3ax)(y-2bx)$

f)

$3xy(a^2+b^2)-ab(x^2+9y^2)=3xya^2+3xyb^2-abx^2-9aby^2$

$=(3xya^2-abx^2)-(9aby^2-3xyb^2)$

$=ax(3ay-bx)-3by(3ay-bx)=(3ay-bx)(ax-3by)$

\(a,ĐK:x\ne-3;x\ne0;y\ne0\\ b,A=\dfrac{1}{x^2\left(x+3\right)+y^2\left(x+3\right)}=\dfrac{1}{\left(x^2+y^2\right)\left(x+3\right)}\\ x=y=0\Leftrightarrow A\in\varnothing\)

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