a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)

À thôi mik tự làm đc rồi ạ !

\(\left|4\frac12x+3\right|-\left|x-1\right|=5\left(x-2\right)\)

=>|4,5x+3|-|x-1|=5(x-2)(1)

TH1: x<-2/3

=>4,5x+3<0; x-1<0

(1) sẽ trở thành:

-4,5x-3-(1-x)=5(x-2)

=>-4,5x-3-1+x=5x-10

=>-3,5x-4=5x-10

=>-8,5x=-10+4=-6

=>\(x=\frac{6}{8,5}=\frac{12}{17}\) (loại)

TH2: -2/3<=x<1

=>4,5x+3>=0; x-1<0

(1) sẽ trở thành:

\(4,5x+3-\left(1-x\right)=5x-10\)

=>4,5x+3-1+x=5x-10

=>5,5x+2=5x-10

=>5,5x-5x=-10-2

=>0,5x=-12

=>x=-24(loại)

TH3: x>=1

(1) sẽ trở thành: 4,5x+3-(x-1)=5x-10

=>4,5x+3-x+1=5x-10

=>3,5x+4=5x-10

=>-1,5x=-14

=>x=14/1,5=28/3(nhận)

a: =>x+3=x-2 hoặc x+3=2-x

=>2x=-1

=>x=-1/2

b: =>3x+7=x-2 hoặc 3x+7=-x+2

=>2x=-9 hoặc 4x=-5

=>x=-5/4 hoặc x=-9/2

c: =>|3x-4|=|2x-5|

=>3x-4=2x-5 hoặc 3x-4=-2x+5

=>x=-1 hoặc x=9/5

1: \(\left(x-1\right)\left(x+5\right)\left(x^2+4x+8\right)+40=0\)

=>\(\left(x^2+4x-5\right)\left(x^2+4x+8\right)+40=0\)

=>\(\left(x^2+4x\right)^2+3\left(x^2+4x\right)-40+40=0\)

=>\(\left(x^2+4x\right)^2+3\left(x^2+4x\right)=0\)

=>\(\left(x^2+4x\right)\left(x^2+4x+3\right)=0\)

=>x(x+4)(x+1)(x+3)=0

=>x∈{0;-4;-1;-3}

2: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-15=0\)

=>\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)-15=0\)

=>\(\left(x^2-5x+4\right)^2+2\left(x^2-5x+4\right)-15=0\)

=>\(\left(x^2-5x+4+5\right)\left(x^2-5x+4-3\right)=0\)

=>\(\left(x^2-5x+9\right)\left(x^2-5x+1\right)=0\)

=>\(x^2-5x+1=0\)

=>\(x^2-5x+\frac{25}{4}-\frac{21}{4}=0\)

=>\(\left(x-\frac52\right)^2=\frac{21}{4}\)

=>\(x-\frac52=\pm\frac{\sqrt{21}}{2}\)

=>\(x=\frac52\pm\frac{\sqrt{21}}{2}\)