Đáp án + Giải thích các bước giải:
`a//`
`ĐK:2x+5\ne0`
`=>2x\ne-5`
`=>x\ne-(5)/(2)`
`b//`
`ĐK:`
$\left\{\begin{matrix}x-4\ne0& \\2x+2\ne0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}x\ne4& \\x\ne-1& \end{matrix}\right.$
Vậy `x\ne{4;-1}`
`c//`
`ĐK:4x^{2}-25\ne0`
`=>(2x+5)(2x-5)\ne0`
`⇒` $\left\{\begin{matrix}2x+5\ne0& \\2x-5\ne0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}2x\ne-5& \\2x\ne5& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}x\ne-\frac{5}{2}& \\x\ne\frac{5}{2}& \end{matrix}\right.$
`⇒x\ne{±(5)/(2)}`
`d//`
`ĐK:` $\left\{\begin{matrix}x-2\ne0⇒x\ne2& \\\frac{2x+3}{x-2}+2\ne0(1)& \end{matrix}\right.$
`(1)<=>(2x+3)/(x-2)+2\ne0`
`⇒(2x+3)/(x-2)\ne-2`
`⇒2x+3\ne-2(x-2)`
`⇒2x+3\ne-2x+4`
`⇒2x+2x\ne-3+4`
`⇒4x\ne1`
`⇒x\ne(1)/(4)`
Vậy `x\ne{(1)/(4);2}`
`e//`
` ĐK:8x^{3}+27\ne0`
`=>(2x+3)(4x^{2}-6x+9)\ne0`
`=>` $\left\{\begin{matrix}2x+3\ne0& \\4x^{2}-6x+9\ne0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}2x\ne-3& \\x^{2}-\frac{6}{4}x+\frac{9}{4}\ne0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x\ne-\frac{3}{2}& \\(x-\frac{3}{4})^{2}+\frac{27}{16}\ne0\text{ (Luôn đúng)}& \end{matrix}\right.$
`⇒x\ne-\frac{3}{2}`
`g//`
`ĐK:(2x+2)(4y^{2}-9)\ne0`
`⇒` $\left\{\begin{matrix}2x+2\ne0& \\4y^{2}-9\ne0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}2x\ne-2& \\(2y-3)(2y+3)\ne0& \end{matrix}\right.$
`⇒` $\left\{\begin{matrix}x\ne-1& \\2y-3\ne0& \\2y+3\ne0\end{matrix}\right.$
`⇒` $\left\{\begin{matrix}x\ne-1& \\y\ne\frac{3}{2}& \\y\ne-\frac{3}{2}\end{matrix}\right.$
Vậy `x\ne-1;y\ne{±(3)/(2)}`
