284:

a: \(\left(x^2-6x+9\right)^2-15\left(x^2-6x+10\right)=1\)

=>\(\left(x^2-6x+9\right)^2-15\left(x^2-6x+9+1\right)-1=0\)

=>\(\left(x^2-6x+9\right)^2-15\left(x^2-6x+9\right)-16=0\)

=>\(\left(x^2-6x+9-16\right)\left(x^2-6x+9+1\right)=0\)

=>\(\left\lbrack\left(x-3\right)^2-16\right\rbrack\left\lbrack\left(x-3\right)^2+1\right\rbrack=0\)

=>\(\left(x-3\right)^2-16=0\)

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

=>(x-7)(x+1)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-1\end{array}\right.\)

b: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

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

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

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34>0\forall x\)

nên \(x^2+2x+1=0\)

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

=>x+1=0

=>x=-1

c: \(\left(x^2-9\right)^2=12x+1\)

=>\(x^4-18x^2+81-12x-1=0\)

=>\(x^4-18x^2-12x+80=0\)

=>\(x^4-6x^3+8x^2+6x^3-36x^2+48x+10x^2-60x+80=0\)

=>\(\left(x^2-6x+8\right)\cdot\left(x^2+6x+10\right)=0\)

mà \(x^2+6x+10=\left(x+3\right)^2+1\ge1>0\forall x\)

nên \(x^2-6x+8=0\)

=>(x-2)(x-4)=0

=>x=2 hoặc x=4

Bài 282:

a: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

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

=>(x-1)(x-4)(x-2)(x-3)=0

=>x∈{1;4;2;3}

b: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)=24\)

=>\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

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

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

=>x∈{-6;1;-4;-1}

c: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

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

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

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

mà \(x^2+x+5=\left(x+\frac12\right)^2+\frac{19}{4}>0\forall x\)

nên \(x^2+x-2=0\)

=>(x+2)(x-1)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=1\end{array}\right.\)

d: \(\left(x^2+x-2\right)\left(x^2+x-3\right)=12\)

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

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

=>\(\left(x^2+x-6\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34>0\forall x\)

nên \(x^2+x-6=0\)

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

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

e: x(x+1)\(\left(x^2+x+1\right)=42\)

=>\(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

=>\(\left(x^2+x\right)^2+\left(x^2+x\right)-42=0\)

=>\(\left(x^2+x+7\right)\left(x^2+x-6\right)=0\)

mà \(x^2+x+7=\left(x+\frac12\right)^2+\frac{27}{4}\ge\frac{27}{4}>0\forall x\)

nên \(x^2+x-6=0\)

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

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)