\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)

\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)

\(=\frac{3x-2}{9x^2-4}\)

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

\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)

\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)

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

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

học tốt

\(P=a^5-a\)

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

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

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

\(=5\left(a-1\right)a\left(a+1\right)+\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)

Nhân thấy   \(5\left(a-1\right)a\left(a+1\right)⋮5\);    \(\left(a-1\right)a\left(a+1\right)⋮3!=6\)

=>   \(5\left(a-1\right)a\left(a+1\right)⋮30\)

                    \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮5!\)

=>   \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮30\)

Vậy P chia hết cho 30

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

Tự cm tiếp

a)  \(A=a^3-b^3-c^3-3abc\)

\(=\left(a-b\right)^3+3ab\left(a-b\right)-c^3-3abc\)

\(=\left(a-b-c\right)\left[\left(a-b\right)^2+c\left(a-b\right)+c^2\right]+3ab\left(a-b-c\right)\)

\(=\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)\)

\(=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)

b)  \(B=a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)+a^2c^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left[\left(a-b\right)+\left(b-c\right)\right]\)

\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left(a-b\right)-a^2c^2\left(b-c\right)\)

\(=a^2\left(a-b\right)\left(b^2-c^2\right)+c^2\left(b-c\right)\left(b^2-a^2\right)\)

\(=a^2\left(a-b\right)\left(b-c\right)\left(b+c\right)+c^2\left(b-c\right)\left(b-a\right)\left(b+a\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2b+a^2c-bc^2-ac^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)

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

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

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

\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)

\(=3-6xy-2+6xy\)

\(=1\)

\(x^2+y^2+z^2=xy+yz+zx\)

<=>   \(2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

<=>  \(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

<=>  \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

<=>  \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\) <=>  \(\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)<=>  \(x=y=z\)  (đpcm)

      \(x^2+y^2+z^2=xy+xz+yz\)

\(\Rightarrow2x^2+2y^2+2z^2=2xy+2xz+2yz\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)(1)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall y;z\\\left(x-z\right)^2\ge0\forall x;z\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z}\) (2)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}\Rightarrow}x=y=z}\)

Chúc bạn học tốt.

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

\(=x^2-4x+4-x^2-6x-9+x^2-16\)

\(=x^2-10x-21\)

do k biết đề nên mk rút gọn

Sửa đề chút: a+b+c=0

Ta có: \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0\)

\(a^2+b^2+c^2+2ab+2bc+2ca=0\)

\(\Rightarrow2\left(ab+bc+ca\right)=0-1\)

\(\Rightarrow ab+bc+ca=-\frac{1}{2}\)

\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2bc^2a+2ca^2b=\frac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc.\left(a+b+c\right)=\frac{1}{4}\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Ta có: \(a^2+b^2+c^2=1\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=1\)

\(a^4+b^4+c^4+2.\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]=1\)

\(a^4+b^4+c^4+2.\frac{1}{4}=1\)

\(a^4+b^4+c^4+\frac{1}{2}=1\)

\(\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)

Vậy \(a^4+b^4+c^4=\frac{1}{2}\)

Nếu mk sửa đề sai thì bảo mk nhé.( mk lm đúng để của b thử rồi nhưng ko ra)

bn sửa đề thì mk cx ra rồi nhưng quan trọng nó là 1 chứ ko phải là 0