Question

Giúp với

Answer 1

n: \(x^2-4x+4=\left(3-5x\right)^2\)

=>\(\left(5x-3\right)^2=\left(x-2\right)^2\)

=>\(\left(5x-3\right)^2-\left(x-2\right)^2=0\)

=>(5x-3-x+2)(5x-3+x-2)=0

=>(4x-1)(6x-5)=0

=>\(\left[\begin{array}{l}4x-1=0\\ 6x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=1\\ 6x=5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac14\\ x=\frac56\end{array}\right.\)

k: \(x^2\left(x^2+2\right)-x^2-2=0\)

=>\(x^2\left(x^2+2\right)-\left(x^2+2\right)=0\)

=>\(\left(x^2+2\right)\left(x^2-1\right)=0\)

=>\(x^2-1=0\)

=>\(x^2=1\)

=>x=1 hoặc x=-1

j: \(x\left(x-6\right)\left(x+6\right)-\left(x-3\right)\left(x^2+3x+9\right)=-9\)

=>\(x\left(x^2-36\right)-\left(x^3-27\right)=-9\)

=>\(x^3-36x-x^3+27=-9\)

=>-36x+27=-9

=>-36x=-36

=>x=1

i: \(x^2+2x-15=0\)

=>\(x^2+5x-3x-15=0\)

=>(x+5)(x-3)=0

=>x=-5 hoặc x=3

h: \(\left(x^2+1\right)\left(3x-4\right)+\left(2x-1\right)\left(2x+1\right)-3x\left(x^2-2\right)=-3\)

=>\(3x^3-4x^2+3x-4+4x^2-1-3x^3+6x=-3\)

=>9x-5=-3

=>9x=2

=>\(x=\frac29\)