Lời giải:
a)
$(3-4x)^2=16(x-3)^2=4^2(x-3)^2=(4x-12)^2$
$\Leftrightarrow [(3-4x)-(4x-12)][(3-4x)+(4x-12)]=0$
$\Leftrightarrow (15-8x)(-9)=0$
$\Rightarrow 15-8x=0\Rightarrow x=\frac{15}{8}$
b)
$(x^2+x+1)^2=(4x-1)^2$
$\Leftrightarrow (x^2+x+1)^2-(4x-1)^2=0$
$\Leftrightarrow (x^2+x+1-4x+1)(x^2+x+1+4x-1)=0$
$\Leftrightarrow (x^2-3x+2)(x^2+5x)=0$
$\Leftrightarrow (x-1)(x-2)x(x+5)=0$
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x=0\\ x+5=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=2\\ x=0\\ x=-5\end{matrix}\right.\)
a) 9 - 24x + 16x2 = 16(x2 - 6x + 9)
=> 16x2 - 24x + 9 = 16x2 - 96x + 144
=> -24x + 96x = 144 - 9
=> 72x = 135
=> x = \(\frac{15}{8}\)
b) (x2 + x + 1)2 = (4x - 1)2
=> x4 + x2 + 1 + 2x3 + 2x + 2x2 = 16x2 - 8x + 1
=> x4 + 2x3 + 3x2 + 2x + 1 = 16x2 - 8x + 1
=> x4 + 2x3 - 13x2 + 10x = 0
=> x4 - x3 + 3x3 - 3x2 - 10x2 + 10x = 0
=> (x3 + 3x2 - 10x)(x - 10) = 0
=> x(x2 + 3x - 10)(x - 10) = 0
=> x(x - 2)(x+5)(x-10) = 0
=> \(\left[{}\begin{matrix}x=0\\x=2\\x=-5\\x=10\end{matrix}\right.\)
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