f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

a) \(x^2+4x+4-y^2\)

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

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

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

\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)

a: \(x^2+4x+4-y^2=\left(x+2-y\right)\left(x+2+y\right)\)

b: \(x^2-4xy+4y^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\)

c: \(x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\)

a: \(x^4+2x^3+x^2=x^2\left(x+1\right)^2\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

a, đặt ( x²+x)=y ta có :

y²+4y=12 <=> y²+4y-12=0

<=> y²+4y+4-16 =0

<=>(y²+4y+4)-16+=0

<=> (y+2)²-16=0

<=>(y-2)(y+6)=0

<=>y-2=0 hoặc y+6=0

<=> y=2 hoặc y=-6

<=> x²+x=2 hoặc x²+x=-6

<=> x²+x -2=0 hoặc x²+x+6=0(vô lý)

<=> (x-1)(x+2)=0 <=> x-1=0 hoặc x+2=0

<=> x=1 hoặc x=-2

vậy pt có nghiệm là x=1 và x=-2

b,6x⁴-5x³-38x²-5x+6=0

<=>6x⁴-18x³+13x³-39x²+x²-3x-2x+6=0

<=>6x³(x-3)+13x²(x-3)+x(x-3)-2(x-3)=0

<=>(x-3)(6x³+13x²+x-2)=0

<=>(x-3)(6x³+12x²+x²+2x-x-2)=0

<=>(x-3)(6x²(x+2)+x(x+2)-(x+2))=0

<=>(x-3)(x+2)(6x²+x-1)=0

<=>(x-3)(x+2)(3x-1)(2x+1)=0

tới đây tự làm

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

a Đề sai: )

b

\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

c: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

d: =x^2(x^2+2x+1)

=x^2(x+1)^2

e: =5(x^2-2xy+y^2-z^2)

=5[(x-y)^2-z^2]

=5(x-y-z)(x-y+z)