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1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
a) A = x2 - 2x + 1 - y2 + 2x - 1
= (x2 - 2x + 1)-( y2-2x+1)
= (x-1)2-(y-1)2
= (x-1-y+1)(x-1+y-1)
b) A = x2 - 4x + 4 - y2 - 6y - 9
= (x2 - 4x + 4)-(y2+6y+9)
= (x-2)2-(y+3)2
= (x-2-y-3)(x-2+y+3)
c) A = 4x2 - 4x + 1 - y2 - 8y - 16
= (4x2 - 4x + 1) - (y2+8y+16)
= (2x-1)2-(y+4)2
= (2x-1-y-4)(2x-1+y+4)
d) A = x2 - 2xy + y2 - z2 + 2zt - t2
=(x2 - 2xy + y2)-(z2- 2zt + t2)
= (x-y)2-(z-t)2
=(x-y-z+t)(z-y+z-t)
câu d mik có sửa lại đề vì mik thấy đề hơi sai
x2 + 4x – 2xy – 4y + y2 = (x2-2xy+ y2) + (4x – 4y) → bạn Việt dùng phương pháp nhóm hạng tử
= (x - y)2 + 4(x – y) → bạn Việt dùng phương pháp dùng hằng đẳng thức và đặt nhân tử chung
= (x – y)(x – y + 4) → bạn Việt dùng phương pháp đặt nhân tử chung
\(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
a) \(x^2+2xy+y^2-4=\left(x+y\right)^2-2^2\)
\(=\left(x+y-2\right)\left(x+y+2\right)\)
b) \(x^2-y^2+x+y=\left(x-y\right)\left(x+y\right)+1\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+1\right)\)
c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)
\(=\left(x+y\right)^2-4^2=\left(x+y+4\right)\left(x+y-4\right)\)
\(\left(x^2+2xy+y^2\right)\left(x^2-2xy+y^2\right)=\left(x-y\right)^2\cdot\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(x^2-2xy-4+y^2=\left(x^2-2xy+y^2\right)-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
\(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)