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\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
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Ta có:\(\left(a-b+c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2\\ =2\left(\left(a-b+c\right)^2-\left(b-c\right)^2\right)\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\\ =2\left(a-2b+2c\right)a \)
\(=2a^2-4ab+4ac\)
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a: =x^2+2xy+y^2-4x^2y^2
=(x+y)^2-(2xy)^2
=(x+y+2xy)(x+y-2xy)
b: =49-(a^2-2ab+b^2)
=49-(a-b)^2
=(7-a+b)(7+a-b)
c: =\(a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left(b-2c\right)^2=\left(a-b+2c\right)\left(a+b-2c\right)\)
d:
\(=\left(bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(bc-b^2-c^2+a^2\right)\left(bc+b^2+c^2-a^2\right)\)
e: \(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-4c^2\)
=2(a+b)^2-2c^2
=2[(a+b)^2-c^2]
=2(a+b-c)(a+b+c)
a(b^2-a^2+a^2-c^2) + b(c^2-a^2) + c(a^2-b^2)
= a(b^2-a^2) + a(a^2-c^2) + b(c^2-a^2) + c(a^2-b^2)
= (a^2-b^2)(c-a) + (c^2-a^2)(b-a)
= (a-b)(a+b)(c-a) + (c-a)(c+a)(b-a)
= (a-b)(c-a)(a+b-c-a) = (a-b)(b-c)(c-a)