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Answer:
\(2x^3+4x^2y+2xy^2\)
\(= 2 x ( x ² + 2 x y + y ² )\)
\(= 2 x ( x + y ) ² \)
\( − 3 x ^4 y − 6 x ^3 y ^2 − 3 x ^2 y ^3 \)
\(=-3x^2y(x^2+2xy+y^2)\)
\(=-3x^2y(x+y)^2\)
\(4x^5y^2+8x^4y^3+4x^3y^4\)
\(=4x^3y^2.x^2+4x^3y^2.2xy+4x^3y^2.y^2\)
\(=4x^3y^2.(x^2+2xy+y^2)\)
\(=4x^3y^2.(x+y)^2\)
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)
a) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-12xy-\frac{12}{5}y^2+12xy=3x^2-\frac{12}{5}y^2\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)
a)\(ĐKXĐ:x\ne0;-1\)
Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)
\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
b)x2+2xy+y2-16=(x+y)2-42=(x+y+4)(x+y-4)
c)3x2+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)
d)4x2-6x3y-2x2+8x=2x(2x-3x2y-x+4)
e)x2-4-2xy+y2=(x2-2xy+y2)-4=(x-y)2-22=(x-y-2)(x-y+2)
k)x2-y2-z2-2yz=x2-(y+z)2=(x-y-z)(x+y+z)
m)6xy+5x-5y-3x2-3y2=3(x2-2xy+y2)+5(x-y)=3(x-y)2+5(x-y)=(x-y)(3x-3y+5)
Lời giải:
\(P=\frac{2xy^2(4x^4y^4+x^2)}{2xy^2}-\frac{3x^3y^2(2x-1)}{3x^3y^2}=4x^4y^4+x^2-(2x-1)\)
\(=4x^4y^4+(x^2-2x+1)=(2x^2y^2)^2+(x-1)^2\)
Do $(2x^2y^2)^2\geq 0; (x-1)^2\geq 0$ với mọi $x,y\in\mathbb{R}$
Do đó $P\geq 0$
Vậy GTNN của $P$ là $0$. Dấu "=" xảy ra khi $2x^2y^2=0$ và $x-1=0$ hay $y=0; x=1$