Question

разложите на множители 3a^2+6ab+3b^2
k^4-81
m^2-n^2-m+n
12a^2b^3+3a^3b^2+16b^2-a^2
срочно

Answer 1

$\tt 3a^2+6ab+3b^2=3(a^2+2ab+b^2)=3(a+b)^2\\\\ k^4-81=(k^2)^2-9^2=(k^2-9)(k^2+9)=(k-3)(k+3)(k^2+9)\\\\ m^2-n^2-m+n=(m-n)(m+n)-(m-n)=(m-n)(m+n-1)\\\\12a^2b^3+3a^3b^2+16b^2-a^2=3a^2b^2(4b+a)+(4b-a)(4b+a)=\\=(4b+a)(3a^2b^2+4b-a)$

Answer 2

$\displaystyle 1). \ \ 3a^{2}+6ab+3b^{2}=3\cdot(a^{2}+2ab+b^{2})=3\cdot(a+b)^{2};\\\\2). \ \ k^{4}-81= (k^{2})^{2}-(3^{2})^{2}=(k^{2}-3^{2})(k^{2}+3^{2})=(k-3)(k+3)(k^{2}+9); \\\\3). \ \ m^{2}-n^{2}-m+n=(m-n)(m+n)-(m-n)=(m-n)(m+n-1);\\\\4). \ \ 12a^{2}b^{3}+3a^{3}b^{2}+16b^{2}-a^{2}=3a^{2}b^{2}(4b+a)+(4b-a)(4b+a)=\\\\{} \ \ =(4b+a)(3a^{2}b^{2}+4b-a);$