\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)

\(=\left(x-y=y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)\(=\left(x-z\right)\left[x^2-2xy+y^2-\left(xy-xz-y^2+yz\right)+y^2-2yz+z^2\right]-\left(x-z\right)^3\)\(=\left(x-z\right)\left(x^2+z^2-3xy-3yz+xz+3y^2\right)-\left(x-z\right)^3\)

\(=\left(x-z\right)\left[\left(x^2+z^2-3xy-3yz+xz+3y^2\right)-\left(x-z\right)^2\right]\)

\(=\left(x-z\right)\left(3y^2-3xy+3xz-3yz\right)\)

\(=3\left(y^2-xy+xz-yz\right)\left(x-z\right)\)

c)

\(=\left(x-1\right)^3+\left(x-2\right)^3+\left(3-2x\right)^3=0\)

\(=\left[\left(x-1\right)+\left(x-2\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(x-2\right)+\left(x-2\right)^2\right]+\left(3-2x\right)^3=0\)

\(=\left(2x-3\right)\left[x^2-2x+1-\left(x^2-3x+2\right)+x^2-4x+4\right]+\left(3-2x\right)^3=0\)

\(=\left(2x-3\right)\left(x^2-3x+3\right)-\left(2x-3\right)^3=0\)

\(=\left(2x-3\right)\left[\left(x^2-3x+3\right)-\left(2x-3\right)^2\right]=0\)

\(=\left(2x-3\right)\left(-3x^2+9x-6\right)=0\)

\(=-3\left(x^2-3x+2\right)\left(2x-3\right)=0\)

\(=-3\left(x-1\right)\left(x-2\right)\left(2x-3\right)=0\)

⇒\(\left[{}\begin{matrix}x-1=0\\x-2=0\\2x-3=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)

bạn cần câu c k ?

b)\(\)

\(x^3-5x^2+8x-4\)

\(=x^3-6x^2+12x-8+x^2-4x+4\)

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

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

a) \(\dfrac{x^2-1}{120}+\dfrac{x^2-2}{119}+\dfrac{x^2-3}{118}=3\)

\(=\dfrac{x^2-1}{120}-1+\dfrac{x^2-2}{119}-1+\dfrac{x^2-3}{118}-1=0\)\(=\dfrac{x^2-121}{120}+\dfrac{x^2-121}{119}+\dfrac{x^2-121}{118}=0\)

\(=\left(x^2-121\right).\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)

\(=\left(x+11\right)\left(x-11\right)\left(\dfrac{1}{120}+\dfrac{1}{119}+\dfrac{1}{118}\right)=0\)

⇒\(\left[{}\begin{matrix}x+11=0\\x-11=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)

\(A=\left(1+2+2^2+...+2^{49}\right)-\left(2^{50}+3\right)\)

\(2A=\left(2+2^3+2^4+...+2^{50}\right)-2\left(2^{50}+3\right)\)

\(2A-A=2+2^3+2^4+...+2^{50}-2\left(2^{50}+3\right)-1-2-2^2-...2^{49}+\left(2^{50}+3\right)\)\(A=2^{50}-3-\left(2^{50}+3\right)\)

\(A=-6\)

\(x^3-2x-4\)

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

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

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

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

Để giá trị của phân thức \(\dfrac{x+6}{x-2}\)được xác định thì:

\(x\)-2< > 0⇒\(x\)< > 2

a)

P=\(\dfrac{x+\dfrac{1}{y}}{\dfrac{1}{x}+y}\)=\(\dfrac{\dfrac{xy+1}{y}}{\dfrac{xy+1}{x}}\)

P=\(\dfrac{xy+1}{y}\)/\(\dfrac{xy+1}{x}\)=\(\dfrac{xy+1}{y}\).\(\dfrac{x}{xy+1}\)

P=\(\dfrac{x}{y}\)

Câu b mk chỉ tìm đc kt qả thôi chứ k bt cách trình bày, bạn thông cảm !