\(x^2+2x+4x+8+n-8:x+2\)

\(n=8\)

hình như đề sai rồi

a,=\(x^4+x^3-3x^2-x^4-3x^2+2x^2+2x+6+4x^2-4x\)\(-8=6-8=-2\)

b,=\(x^2-x-6+x^2-1-2x^2+x=-7\)

g,=\(\left(x^3-27y^3\right)\left(2x+y\right)\)\(=2x^4+x^3y-54xy^3-27y^4\)

h,=\(x^6+y^9\)

=cosy+\(\dfrac{sin^2y}{cosy}\)=\(\dfrac{1}{cosy}\)

a,(a+b+c-c)²=a²

b,(a+b+c)²+2a²+2b²+2c²-2ab-2ac-2bc-3(a²+b²+c²)=3a²+3b²+3c²-3a²-3b²-3c²=0

=(1-1).\(\dfrac{189}{2007}\)=0

=3\(x^2\)+2\(x^2\)+8\(x\)+6-5\(x^2\)+10\(x\)=18\(x\)+6

27<\(3^x\)<243

3<\(x\)<5

\(x\)=4

\(\dfrac{1}{x-3}\)=\(\dfrac{x^2-3x+5}{x^2-x-6}\)

suy ra \(x\)²-3\(x\)+5=\(x\)=2