3a^2 - 6ab + 3b^2 - 12c^2
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\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left(\left(a-b\right)^2-\left(2c\right)^2\right)\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
a) 3a2-6ab+3b2-12c2
=3.(a2-2ab+b2-4c2)
=3.[(a-b)2-4c2]
=3.(a-b-2c)(a-b+2c)
\(3a^2-6ab-3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=\left(\sqrt{3}\left(a-b\right)\right)^2-\left(\sqrt{12}c\right)^2=\left(\sqrt{3}a-\sqrt{3}b-\sqrt{12}c\right)\left(\sqrt{3}a-\sqrt{3}b+\sqrt{12}c\right)\)
Sửa đề :\(3a^2-6ab+3b^2-12c^2\)=\(3\left(a^2-2ab+b^2-4c^2\right)\)
=\(3\left(a-b\right)^2-3\left(2c\right)^2\)=\(3\left(a-b-c\right)\left(a-b+c\right)\)
3a2 - 6ab + 3b2 - 12c2
= 3a2 - 3a x 2b + 3b2 - 3 x 4c2
= 3(a2 -a x b +b2 - 4c2)
3a^2 -6ab +3ab^2 -12c^2
= 3(a^2 -2ab +b^2 -4c^2)
= 3[(a- b)^2 -4c^2]=3(a- b+ 4c)(a- b- 4c)
3a2 - 6ab + 3b2 - 12c2
= 3a2 - 3.2ab + 3b2 - 3.4c2
= 3 (a2 - 2ab + b2 - 4c2)
\(3a^2-6ab+3b^2-12c^2\)
\(=3a^2-3ab-3ab+3b^2-12c^2\)
\(=\left(3a^2-3ab\right)-\left(3ab-3b^2\right)-12c^2\)
\(=3a\left(a-b\right)-3b\left(a-b\right)-12c^2\)
\(=\left(3a-3b\right)\left(a-b\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
b) a2 + 2ab + b2 - ac - bc
= ( a + b)2 - c (a + b)
= ( a + b).( a + b - c )
c) x3 - x2 - x + 1
= x2 .(x - 1) - ( x - 1)
= ( x- 1).( x2 - 1)
= ( x-1).(x-1).(x+1)
1)\(3a^2+6ab+3b^2-12c^2\)
\(=3\left(a^2+2ab+b^2-4c^2\right)\)
\(=3\left[\left(a^2+2ab+b^2\right)-4c^2\right]\)
\(=3\left[\left(a+b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a+b-2c\right)\left(a+b+2c\right)\)
3a2 - 6ab + 3b2 - 12c2 = 3.(a2 - 2ab + b2 - 4c2) = 3.[(a - b)2 - 4c2] = 3.(a - b - 2c).(a - b + 2c)