Bài 1 :
\(\dfrac{3x+2}{x-5}\) xác định \(\Leftrightarrow x-5\ne0\Leftrightarrow x\ne5\)
\(\dfrac{x+15}{x-2y}\) xác định \(\Leftrightarrow x-2y\ne0\Leftrightarrow x\ne2y\)
\(\dfrac{x+3}{x^2+1}\) xác định \(\Leftrightarrow x^2+1\ne0\Leftrightarrow\forall x\in R\)
\(\dfrac{5x-4}{\left(3x-2\right)\left(3x+2\right)}\) xác định \(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}3x-2\ne0\\3x+2\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{2}{3}\\x\ne-\dfrac{2}{3}\end{matrix}\right.\)
1:
a: ĐKXĐ: x-5<>0
=>x<>5
b: ĐKXĐ: x-2y<>0
=>x<>2y
c: ĐKXĐ: x^2+1<>0
=>x^2<>-1(luôn đúng)
d: ĐKXĐ: (3x-2)(3x+2)<>0
=>3x-2<>0 và 3x+2<>0
=>x<>2/3 và x<>-2/3
2:
a: ĐKXĐ: x<>-1
\(A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)^2}=\dfrac{3x}{x+1}\)
Khi x=-2 thì \(A=\dfrac{3\cdot\left(-2\right)}{-2+1}=6\)
b: ĐKXĐ: x<>y; x<>-y
\(B=\dfrac{y\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x+y}\)
Khi x=4 và y=-2 thì \(B=\dfrac{-2}{-2+4}=-\dfrac{2}{2}=-1\)
Bài 2 :
\(A=\dfrac{3x^2+3x}{x^2+2x+1}\)
\(\Leftrightarrow A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)^2}\left(x\ne-1\right)\)
\(\Leftrightarrow A=\dfrac{3x}{x+1}\)
\(\Leftrightarrow A=\dfrac{3.\left(-2\right)}{-2+1}=6\left(x=-2\right)\)
\(B=\dfrac{xy-y^2}{x^2-y^2}\)
\(\Leftrightarrow B=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\left(x\ne\pm y\right)\)
\(\Leftrightarrow B=\dfrac{x}{\left(x+y\right)}\)
\(\Leftrightarrow B=\dfrac{4}{4-2}=2\left(x=4;y=-2\right)\)