(2^x)² - 5·2^x·2² +64 > 0
2^x = z
z² -20 z +64 > 0 ищем корни z1 = 16 и z2 = 4
2^x < 4 и z^x > 16
x< 2 x > 4
<span>а)(1+3m)(1-3m)=1-9m^2</span>
<span><span>в)(2x-y)(2x+y)=4x^2-y</span></span>
<span><span><span>д)(4x+3y)(3y-4x)=-(4x+3y)(4x-3y)=-16x^2+9y^2</span></span></span>
<span><span><span><span>номер2</span></span></span></span>
<span><span><span><span><span>а)(x(^2)+2)(x(^2)-2)=x^4-4</span></span></span></span></span>
<span><span><span><span><span><span>в)(a(^2)-4)(a(^2)+4)=a^4-16</span></span></span></span></span></span>
<span><span><span><span><span><span><span>д)(ab-c)(ab+c)=a^2b^2-c^2</span></span></span></span></span></span></span>
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