Ответ:
раскрываем скобки по формуле a^2-2ab+b^2
[tex]\displaystyle\bf\\1)\\\\61^{2} -60^{2}=(61-60)\cdot(61+60)=1\cdot 121=121\\\\51^{2} -50^{2}=(51-50)\cdot(51+50)=1\cdot 101=101\\\\121 > 101 \ \ \ \Rightarrow \ \ \ 61^{2} -60^{2} > 51^{2} - 50^{2} \\\\\\2)\\\\x^{2} -0,04\\\\(x-0,2)(x+0,2)=x^{2} -0,2^{2} =x^{2} -0,04\\\\x^{2} -0,04=(x-0,2)(x+0,2)\\\\\\3)\\\\(x-3)^{2} =x^{2} -6x+9=(x^{2} -6x+8)+1\\\\x^{2} -6x+8\\\\(x-3)^{2} > x^{2} -6x+8[/tex]
[tex]\displaystyle\bf\\4)\\\\x^{2} +4x+3=(x^{2} +4x+4)-1\\\\(x+2)^{2} =x^{2} +4x+4\\\\x^{2} +4x+3 < (x+2)^{2} \\\\\\5)\\\\(x-2)(x+2)=x^{2} -4=(x^{2} -1)-3\\\\(x+1)(x-1)=x^{2} -1\\\\(x-2)(x+2) < (x+1)(x-1)\\\\\\6)\\\\(0,5a^{2} b+2b)(0,5a^{2}b-2b) +(0,5a^{2}b+2b)^{2} =\\\\=(0,5a^{2} b)^{2} -(2b)^{2} +(0,5a^{2} b)^{2} +2\cdot 0,5a^{2} b\cdot 2b+(2b)^{2} =\\\\=0,25a^{4} b^{2} -4b^{2} +0,25a^{4}b^{2} +2a^{2} b^{2} +4b^{2}=\boxed{0,5a^{4} b^{2}+2a^{2} b^{2}}[/tex]
[tex]\displaystyle\bf\\7)\\\\(1+2x)(2x-1)-\frac{1}{4} (4x-8)^{2} =(2x)^{2} -1^{2} -\frac{1}{4} \cdot 16\cdot(x-2)^{2} =\\\\\\=4x^{2} -1-4\cdot(x^{2} -4x+4)=4x^{2} -1-4x^{2} +16x-16=\boxed{16x-17}\\\\\\8)\\\\(8-n^{3} )(8+n^{3})=8^{2} -(n^{3} )^{2} =\boxed{64-n^{6}}[/tex]
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Ответ:
раскрываем скобки по формуле a^2-2ab+b^2
Verified answer
[tex]\displaystyle\bf\\1)\\\\61^{2} -60^{2}=(61-60)\cdot(61+60)=1\cdot 121=121\\\\51^{2} -50^{2}=(51-50)\cdot(51+50)=1\cdot 101=101\\\\121 > 101 \ \ \ \Rightarrow \ \ \ 61^{2} -60^{2} > 51^{2} - 50^{2} \\\\\\2)\\\\x^{2} -0,04\\\\(x-0,2)(x+0,2)=x^{2} -0,2^{2} =x^{2} -0,04\\\\x^{2} -0,04=(x-0,2)(x+0,2)\\\\\\3)\\\\(x-3)^{2} =x^{2} -6x+9=(x^{2} -6x+8)+1\\\\x^{2} -6x+8\\\\(x-3)^{2} > x^{2} -6x+8[/tex]
[tex]\displaystyle\bf\\4)\\\\x^{2} +4x+3=(x^{2} +4x+4)-1\\\\(x+2)^{2} =x^{2} +4x+4\\\\x^{2} +4x+3 < (x+2)^{2} \\\\\\5)\\\\(x-2)(x+2)=x^{2} -4=(x^{2} -1)-3\\\\(x+1)(x-1)=x^{2} -1\\\\(x-2)(x+2) < (x+1)(x-1)\\\\\\6)\\\\(0,5a^{2} b+2b)(0,5a^{2}b-2b) +(0,5a^{2}b+2b)^{2} =\\\\=(0,5a^{2} b)^{2} -(2b)^{2} +(0,5a^{2} b)^{2} +2\cdot 0,5a^{2} b\cdot 2b+(2b)^{2} =\\\\=0,25a^{4} b^{2} -4b^{2} +0,25a^{4}b^{2} +2a^{2} b^{2} +4b^{2}=\boxed{0,5a^{4} b^{2}+2a^{2} b^{2}}[/tex]
[tex]\displaystyle\bf\\7)\\\\(1+2x)(2x-1)-\frac{1}{4} (4x-8)^{2} =(2x)^{2} -1^{2} -\frac{1}{4} \cdot 16\cdot(x-2)^{2} =\\\\\\=4x^{2} -1-4\cdot(x^{2} -4x+4)=4x^{2} -1-4x^{2} +16x-16=\boxed{16x-17}\\\\\\8)\\\\(8-n^{3} )(8+n^{3})=8^{2} -(n^{3} )^{2} =\boxed{64-n^{6}}[/tex]