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Question 1

Question 2

Решите задачу:

$1) x^{2} =9\\x=+-3 \\ \\ 2)( x^{2} -2x+1) ^{2} =1^{2} \\ ( x^{2} -2x+1) ^{2} -1^{2} =0\\( x^{2} -2x+1-1=0)( x^{2} -2x+1+1)=0 \\ x^{2} -2x=0\\x(x-2)=0\\x=0\\x-2=0\\x=0\\$

$x^{2} -2x+2=0\\D=4+8=12 \\ \sqrt{D} = \sqrt{3*4} =2 \sqrt{3} \\ x_{1} = \frac{2+2 \sqrt{3} }{2} = \frac{2(1+ \sqrt{3} )}{2} =1+ \sqrt{3} \\ \\ x_{1} = \frac{2-2 \sqrt{3} }{2} = \frac{2(1- \sqrt{3} )}{2} =1- \sqrt{3}$

$3)(x+1)^{2} =(2x+5)^{2} \\ x^{2} +1+2x=4 x^{2} +25+20x \\ x^{2} +1+2x-4 x^{2} -25-20x=0\\-3 x^{2} -18x-24=0/*(-1)\\3 x^{2} +18x+24=0/:3\\ x^{2} +6x+8=0\\D=36-32=4 \\ \sqrt{D} =2 \\ x_{1} = \frac{-6-2}{2} = \frac{-8}{2} =-4 \\ \\ x_{2} = \frac{-6+2}{2} = \frac{-4}{2} =-2$

$4)2 x^{2} -7x+5=0\\D=49-40=9 \\ \sqrt{D} =3 \\ x_{1} = \frac{7+3}{4} = \frac{10}{4} =2.5 \\ x_{2} = \frac{7-3}{4} = \frac{4}{4} =1$

$5)3 x^{2} -7x+5=0\\D=49-60<0$

$6) x^{2} -2011x+2010=0 \\ D=2011^{2} -4*2010=2009^{2} \\ \sqrt{D} =2009 \\ x_{1} = \frac{2011-2009}{2} = \frac{2}{2} =1 \\ x_{2} = \frac{2011+2009}{2} = \frac{4020}{2} =2010$

$7) x^{2} -2010x-2011=0\\D=(-2010 )^{2} +4*2011=2012^{2} \\ \sqrt{D} =2012 \\ x_{1} = \frac{2010-2012}{2} =-1 \\ x_{2} = \frac{2010+2012}{2} =2011 \\$

$8)2 x^{4} -7 x^{2} +5=0\\ x^{2} =t\\2t ^{2} -7t+5=0\\D=49-40=9 \\ \sqrt{D} =3\\t_{1} = \frac{7+3}{4} = \frac{10}{4} =2.5 \\ t _{2} = \frac{7-3}{4} =1 \\ x^{2} =2.5\\x=+- \sqrt{2.5} \\ x^{2} =1\\x=+-1$