Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=x+\sqrt{x}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

\(=x+\sqrt{x}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

Ta có : \(x^2+3x+2=x^2-2x+1+5x-5+6\)

\(=\left(x-1\right)^2+5\left(x-1\right)+6\)

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

Đề hơi sai sai nhỉ :))

cảm ơn

giúp mik với

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

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

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

\(=2x^3-3x-1\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2022}\\ \Leftrightarrow2A-A=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}\\ \Leftrightarrow A=2^{2022}-1\\ \Leftrightarrow A+1=2^{2022}\)

Mà \(A+1=2^x\Leftrightarrow x=2022\)

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

b) \(\dfrac{x^2+2xy+y^2}{2x^2+xy-y^2}\)

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

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

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=ax\left(x-a\right)\)

..

đc ko bạn

cảm ơn bạn nha

\(9x^2-6x+1-9x^2-9x=-15x+1\)

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

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

=-15x+1