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

\(=8x^3-12x^2+6x-1-\left(3x^3-6x^2-x+2\right)+x^3+9x^2+27x+27\)

\(=9x^3-3x^2+33x+26-3x^3+6x^2+x-2\)

\(=6x^3+3x^2+34x+24\)

a: ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

b: \(B=\dfrac{x+3-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+2+1}{x+2}\)

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

c: Để B nguyên thì \(x-3\in\left\{1;-1\right\}\)

hay \(x\in\left\{4;2\right\}\)

Lời giải:

$(x-1)^3-(x-1)(x^2+x+1)=(x-1)[(x-1)^2-(x^2+x+1)]=(x-1)(x^2-2x+1-x^2-x-1)=(x-1)(-3x)=-3x(x-1)$

\(:P=5x(x-3)(x+3)-(2x-3)^2+34x(x+2)-5(x+2)^3+25x-1\)

\(P=5x(x^2-9)-(4x^2-12x+9)+34x^2+68x-5(x^3+6x^2+12x+8)+25-1\)

\(P=5x^3-45x-4x^2+12x-9+34x^2+68x-5x^3-30x^2-60x-40+25-1\)

\(P=(5x^3-5x^3)+(34x^2-4x^2-30x^2)+(12x-45x++68x+25x-60x)-(9+1)\)

\(P=-10\)

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

Dễ thấy 5=4+1=x+1

Thay vào C,ta có:

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+3=x+1\\2x+3=-x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=1-3\\2x+x=-1-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\3x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}\)

-x+1 nha

giup minh voi cac ban