$\text{a)}$  

${\left(x-1\right)}^2=\left(x+1\right)\left(x-1\right)$  

$x^2-2\cdot x\cdot 1+1^2=x^2-1^2$  

$x^2-2x+1=x^2-1\mid -x^2$  

$-2x+1=-1\mid -1$  

$-2x=-2\mid :\left(-2\right)$  

$x=1$  

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$\text{b)}$  

${\left(2x+1\right)}^2=4\left(x+2\right)\left(x-2\right)$  

${\left(2x\right)}^2+2\cdot 2x\cdot 1+1^2=4\left(x^2-2^2\right)$  

$4x^2+4x+1=4x^2-16\mid -4x^2$  

$4x+1=-16\mid -1$  

$4x=-17\mid :4$  

$x=-\frac{17}{4}$  

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$\text{c)}$  

$9{\left(1-2x\right)}^2-4{\left(3x-1\right)}^2=1$  

${\left(3\left(1-2x\right)\right)}^2-{\left(2\left(3x-1\right)\right)}^2=1$  

$\left[3\left(1-2x\right)-2\left(3x-1\right)\right]\cdot \left[3\left(1-2x\right)+2\left(3x-1\right)\right]=1$  

$\left[3-6x-6x+2\right]\cdot \left[3-6x+6x-2\right]=1$  

$\left(5-12x\right)\cdot 1=1$  

$5-12x=1\mid -5$  

$-12x=-4\mid :\left(-12\right)$  

$x=\frac{4}{12}=\frac{1}{3}$  

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$\text{d)}$  

$\left(3x-2\right)\left(3x+2\right)=9{\left(x-\frac{1}{3}\right)}^2$  

${\left(3x\right)}^2-2^2=9\left(x^2-2\cdot x\cdot \frac{1}{3}+{\left(\frac{1}{3}\right)}^2\right)$  

$9x^2-4=9\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)$  

$9x^2-4=9x^2-6x+1\mid -9x^2$  

$-4=-6x+1\mid +6x$  

$6x-4=1\mid +4$  

$6x=5\mid :6$  

$x=\frac{5}{6}$