\(a,=\frac{x^3+26x-19}{\left(x-3\right)\left(x+1\right)}+\frac{2x}{x+1}-\frac{x+3}{x-3}\)

\(=\frac{x^3+26x-19+2x^2-6x-x^2-4x-4}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^3+x^2+16x-23}{\left(x-3\right)\left(x+1\right)}\)

e hocjccccccccccccccccccccc lớp 44444444444444444444444444

\(x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3\)

\(4x^3+32x^2+64x=4x\left(x^2+8x+16\right)=4x\left(x+4\right)^2\)

\(\frac{4x}{\left(x+4\right)^3}=\frac{16x^2}{4x\left(x+4\right)^3},\frac{x-4}{4x\left(x+4\right)^2}=\frac{x^2-16}{4x\left(x+4\right)^3}\)

đề bài là j?

ờ đúng đấy. , không có đề bài

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=8\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=8\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=8\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-1^2-8=0\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-3^2=0\)

\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x+2=0\\x^2+5x+8=0\end{cases}}\Leftrightarrow x=\frac{-5\pm\sqrt{17}}{2}\).

\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow\frac{xy}{ab}=\frac{yz}{bc}=\frac{xz}{ac}=\frac{xy+yz+xz}{ab+bc+ac}.\)(1)

Ta có

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

\(\Leftrightarrow1=1+2\left(ab+bc+ac\right)\Rightarrow ab+bc+ac=0\) => (1) vô nghĩa bạn xem lại đề bài

ta có 

a. \(ab+bc+ac\le a^2+b^2+c^2\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) luôn đúng

mà ta lại có : 

\(\hept{\begin{cases}a^2< a\left(b+c\right)\\b^2< b\left(a+c\right)\\c^2< c\left(a+b\right)\end{cases}\Rightarrow a^2+b^2+c^2\le2\left(ab+bc+ac\right)}\) vậy ta có điều phải chứng minh.

b. ta có  :

\(\hept{\begin{cases}\left(a+b-c\right)\left(a+c-b\right)\le\left(\frac{a+b-c+a+c-b}{2}\right)^2=a^2\\\left(a+b-c\right)\left(b+c-a\right)\le b^2\\\left(a+c-b\right)\left(b+c-a\right)\le c^2\end{cases}}\)

nhân lại ta sẽ có : \(\left(a+b-c\right)\left(a+c-b\right)\left(b+c-a\right)\le abc\) vậy ta có dpcm