ai giải gimf nha

phần này học từ kì 1 rồi mà bn

co cau hoi nao dao ma bao giai

\(=\left(3x+15\right)^2-\left(x-7\right)^2\\ =\left(3x+15-x+7\right)\left(3x+15+x-7\right)\\ =\left(2x+22\right)\left(4x+8\right)\\ =2\left(x+11\right)4\left(x+2\right)\\ =8\left(x+11\right)\left(x+2\right)\)

\(=\left(x^2+y^2-5\right)^2-\left[2\left(xy+2\right)\right]^2\\ =\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\\ =\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\\ =\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)

\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)

\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)

\(=\left(x^2+y^2-2xy-9\right)\left(x^2+y^2+2xy-1\right)\)

\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)

\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)

\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)

\(=\left[\left(x-y\right)^2-9\right]\left[\left(x-y\right)^2-1\right]\)

\(=\left(x-y-3\right)\left(x-y+3\right)\left(x-y-1\right)\left(x-y+1\right)\)

\(-\left(x+2y\right)^2\)

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)

\(3xy-21x^2y^2=3xy\left(1-7xy\right)\)

\(3x^2y-5xy^2+7xy\)

\(=xy\left(3x-5y+7\right)\)

\(3xy-21x^2y^2\\ =3xy-3xy.7xy\\ =3xy\left(1-7xy\right)\)

Cái thứ hai không phân tích bằng đặt nhân tử chung được đâu nha=)

\(3x-6y\\ =3\left(x-2y\right)\)

\(=\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{6}-3-\sqrt{6}\)

=căn 3-2căn 6-3

\(x^3-4x\)

\(=x\cdot\left(x^2-4\right)\)

\(=x\cdot\left(x^2-2^2\right)\)

\(=x\left(x+2\right)\left(x-2\right)\)

\(x^3-4x\)

\(=x\left(x^2-4\right)\)

\(=x\left(x-2\right)\left(x+2\right)\)

#Ayumu

85.12,7+5.3.12,7=12,7.(85+5.3)=12,7.(85+15)=12,7.100=1270