\(1.\)

\(a.\)

\(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)

\(=\left(x^3-3^3\right)-\left(54+x^3\right)\)

\(=x^3-27-54-x^3\)

\(=-81\)

\(b.\)

\(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

\(=\left(27x^3+y^3\right)-\left(27x^3-y^3\right)\)

\(=27x^3+y^3-27x^3+y^3\)

\(=2y^3\)

\(2.\)

\(a.\)

\(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

\(b.\)

\(\left(2x-3y\right)\left(4x^2+6xy+9y^3\right)=8x^3-27y^3\)

1) a) \(\left(x-3\right)\left(x^2+3x+9\right)-\left(54+x^3\right)\)

\(=\left(x^3-3^3\right)-\left(54+x^3\right)\\ =\left(x^3-27\right)-54-x^3\\ =-27-54\\ =-81\)

b) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

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

2) a) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

b) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=8x^3-27y^3\)

\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}\)

\(=\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)

\(=-\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}\)

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

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

a) \(3x^4y-12x^2y^3=3x^2y\left(x^2-\left(2y\right)^2\right)=3x^2y\left(x+2y\right)\left(x-2y\right)\)

b) Sửa đề: \(x^2-y^2-8x+16=\left(x-4\right)^2-y^2=\left(x-4-y\right)\left(x-4+y\right)\)

c) \(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

d) \(3x^2-6xy+3y^2-27=3\left(x^2-2xy+y^2-9\right)=3\left(\left(x-y^2\right)-3^2\right)=3\left(x-y-3\right)\left(x-y+3\right)\) 

\(3x^4y-12x^2y^3=3x^2y\left(x^2-4y^2\right)=3x^2y\left(x-2y\right)\left(x+2y\right)\)

\(x^2-y^2-8y-16=x^2-\left(y^2+8y+16\right)=x^2-\left(y+4\right)^2=\left(x+y+4\right)\left(x-y-4\right)\)

\(x^3+3x^2+4x+12=x^2\left(x+3\right)+4\left(x+3\right)=\left(x^2+4\right)\left(x+3\right)\)

\(3x^2-6xy+3y^2-27=3\left[\left(x-y\right)^2-9\right]=3\left(x-y-3\right)\left(x-y+3\right)\)

Lời giải:

ĐKXĐ: \(x\geq 0; x\neq 1\)

Ta có:

\(A=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}=\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}+\frac{\sqrt{x}-1}{(\sqrt{x}+2)(\sqrt{x}-1)}\)

\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{x+3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

ĐKXĐ : x khác -1 và 1

A = [x^3+1-(x^2-1).(x+1)/(x-1).(x+1)] : [x.(x-1)+x/x-1]

 = [-x^2+x/(x-1).(x+1)] : x^2/x-1

 = -x.(x-1)/(x-1).(x+1) . (x-1)/x^2

 = -(x-1)/x.(x+1)

k mk nha

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