Помогите!!!))))))))))

Question 1

Question 2

$1.\\1)\;(p+8)^2=p^2+16p+64\\2)\;(10x-3y)^2=100x^2-60xy+9y^2\\3)\;(x-9)(x+9)=x^2-81\\(4m+7n)(7n-4m)=49n^2-16m^2\\\\2.\\1)\;16-c^2=(4-c)(4+c)\\2)\;p^2+2p+1=(p+1)^2\\3)\;9m^2-25=(3m-5)(3m+5)\\4)\;36m^2+24mn+4n^2=(6m+2n)^2\\\\3.\;(a-10)^2-(a-5)(a+5)=a^2-20a+100-(a^2-25)=\\=a^2-20a+100-a^2+25=125-20a$

0\;npu\;Bcex\;x\\\\6.\;(2-x)(2+x)(4+x^2)+(6-x)^2=(4-x^2)(4+x^2)+36-12x+x^2=\\=16-x^4+36-12x+x^2=-x^4+x^2-12x+52" alt="4.\;(2x-7)(x+1)+3(4x-1)(4x+1)=2(5x-2)^2-53\\2x^2-5x-7+3(16x^2-1)=2(25x^2-20x+4)-53\\2x^2-5x-7+48x^2-3=50x^2-40x+8-53\\50x^2-5x-10=50x^2-40x-45\\35x=-35\\x=-1\\\\5.\;x^2-18x+84=(x^2-18x+81)+3=(x-9)^2+3>0\;npu\;Bcex\;x\\\\6.\;(2-x)(2+x)(4+x^2)+(6-x)^2=(4-x^2)(4+x^2)+36-12x+x^2=\\=16-x^4+36-12x+x^2=-x^4+x^2-12x+52" align="absmiddle" class="latex-formula">

Trover_zn · Answer 1 · 2020-05-28T17:44:21+0000

$1.\\1)\;(p+8)^2=p^2+16p+64\\2)\;(10x-3y)^2=100x^2-60xy+9y^2\\3)\;(x-9)(x+9)=x^2-81\\(4m+7n)(7n-4m)=49n^2-16m^2\\\\2.\\1)\;16-c^2=(4-c)(4+c)\\2)\;p^2+2p+1=(p+1)^2\\3)\;9m^2-25=(3m-5)(3m+5)\\4)\;36m^2+24mn+4n^2=(6m+2n)^2\\\\3.\;(a-10)^2-(a-5)(a+5)=a^2-20a+100-(a^2-25)=\\=a^2-20a+100-a^2+25=125-20a$

0\;npu\;Bcex\;x\\\\6.\;(2-x)(2+x)(4+x^2)+(6-x)^2=(4-x^2)(4+x^2)+36-12x+x^2=\\=16-x^4+36-12x+x^2=-x^4+x^2-12x+52" alt="4.\;(2x-7)(x+1)+3(4x-1)(4x+1)=2(5x-2)^2-53\\2x^2-5x-7+3(16x^2-1)=2(25x^2-20x+4)-53\\2x^2-5x-7+48x^2-3=50x^2-40x+8-53\\50x^2-5x-10=50x^2-40x-45\\35x=-35\\x=-1\\\\5.\;x^2-18x+84=(x^2-18x+81)+3=(x-9)^2+3>0\;npu\;Bcex\;x\\\\6.\;(2-x)(2+x)(4+x^2)+(6-x)^2=(4-x^2)(4+x^2)+36-12x+x^2=\\=16-x^4+36-12x+x^2=-x^4+x^2-12x+52" align="absmiddle" class="latex-formula">