ĐK: x#0; x#-1

\(\frac{x^4}{1-x}\)+ x³ + x^{2 + 1}

^{= \(\frac{x^4}{1-x}\)}+ \(\frac{x^3\left(1-x\right)}{1-x}\)+ \(\frac{x^2\left(1-x\right)}{1-x}\)+ \(\frac{1-x}{1-x}\)

= \(\frac{x^4+x^3-x^4+x^2-x^3+1-x}{1-x}\)

= \(\frac{x+1}{1-x}\)

Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)

\(=\left(5x^4+4x^4\right)+\left(-3x^3+3x^3\right)+\left(x^2+x^2\right)-x+5\)

\(=9x^4+2x^2-x+5\)

\(Q\left(x\right)=x-5x_{}^3-x^2-x^4+4x^3-x^2+3x-1\)

\(=\left(-5x^4-x^4\right)+4x^3+\left(-x^2-x^2\right)+\left(x+3x\right)-1\)

\(=-6x^4+4x^3-2x^2+4x-1\)

P(x)+Q(x)

\(=9x^4+2x^2-x+5-6x^4+4x^3-2x^2+4x-1\)

\(=3x^4+4x^3+3x+4\)

Đặt P(x)+Q(x)=0

=>\(3x^4+4x^3+3x+4=0\)

=>\(x^3\left(3x+4\right)+\left(3x+4\right)=0\)

=>\(\left(3x+4\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}3x+4=0\\ x^3+1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=-\frac43\\ x=-1\end{array}\right.\)

B1 :

a) (2x - 1)²

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a,$\frac{5}{2x^2y}+\frac{3}{5xy^2}+\frac{x}{y^3}$52x2y +35xy2 +xy3 

b,\(\frac{x+1}{2x+6}+\frac{2x+3}{x\left(x+3\right)}\)

Cộng vào sẽ ra kết quả nha !!! 

We are one_I Love you làm j ngu người quá 

\(x^5+x^4+x^3+x^2+x+1\)

\(=\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)

Ta có: \(x^5+x^4+x^3+x^2+x+1\)

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

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

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

\(=\left(x+1\right)\left\lbrack\left(x^2+1\right)^2-x^2\right\rbrack=\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)