a: ĐKXĐ: \(x^2+3x-10<>0\)
=>(x+5)(x-2)<>0
=>x∉{-5;2}
\(\frac{x^2-4}{x^2+3x-10}=0\)
=>\(x^2-4=0\)
=>\(x^2=4\)
=>\(\left[\begin{array}{l}x=2\left(loại\right)\\ x=-2\left(nhận\right)\end{array}\right.\)
b: ĐKXĐ: \(x^3-3x^2-4x<>0\)
=>\(x\left(x^2-3x-4\right)<>0\)
=>x(x-4)(x+1)<>0
=>x∉{0;4;-1}
TA có: \(\frac{x^3-16x}{x^3-3x^2-4x}=0\)
=>\(\frac{x\left(x^2-16\right)}{x\left(x^2-3x-4\right)}=0\)
=>\(\frac{x^2-16}{x^2-3x-4}=0\)
=>\(\frac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+1\right)}=0\)
=>\(\frac{x+4}{x+1}=0\)
=>x+4=0
=>x=-4(nhận)
c: ĐKXĐ: \(x^3+2x-3<>0\)
=>\(x^3-x+3x-3<>0\)
=>\(\left(x-1\right)\left(x^2+x+3\right)<>0\)
mà \(x^2+x+3=x^2+x+\frac14+\frac{11}{4}=\left(x+\frac12\right)^2+\frac{11}{4}<>0\forall x\)
nên x-1<>0
=>x<>1
\(\frac{x^3+x^2-x-1}{x^3+2x-3}=0\)
=>\(x^3+x^2-x-1=0\)
=>\(x^2\left(x+1\right)-\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2-1\right)=0\)
=>\(\left(x+1\right)^2\cdot\left(x-1\right)=0\)
=>\(\left[\begin{array}{l}x=1\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)
