LN
NT

a: \(x^3-25x=0\)

=>\(x\left(x^2-25\right)=0\)

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

=>\(\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

b: \(x^4+4=5x^2\)

=>\(x^4-5x^2+4=0\)

=>\(\left(x^2-1\right)\left(x^2-4\right)=0\)

=>\(\left[{}\begin{matrix}x^2=1\\x^2=4\end{matrix}\right.\Leftrightarrow x\in\left\{1;-1;2;-2\right\}\)

c: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

=>\(\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

d: \(4\left(x-2\right)^2=25\left(1-2x\right)^2\)

=>\(\left(10x-5\right)^2=\left(2x-4\right)^2\)

=>\(\left(10x-5-2x+4\right)\left(10x-5+2x-4\right)=0\)

=>(8x-1)(12x-9)=0

=>\(\left[{}\begin{matrix}8x-1=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{4}\end{matrix}\right.\)

e: \(\left(3x-5\right)\left(2x-1\right)-\left(x+2\right)\left(6x-1\right)=0\)

=>\(6x^2-3x-10x+5-\left(6x^2-x+12x-2\right)=0\)

=>\(6x^2-13x+5-6x^2-11x+2=0\)

=>-24x+7=0

=>-24x=-7

=>\(x=\dfrac{7}{24}\)

f: \(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=5\)

=>\(9x^2-4-\left(9x^2-6x+1\right)=5\)

=>\(9x^2-4-9x^2+6x-1=5\)

=>6x-5=5

=>6x=10

=>\(x=\dfrac{5}{3}\)

Bình luận (0)