Xét tam giác ABE vuông tại E và tam giác ACE vuông tại E

Có : EB=EC(E là trung điểm BC)

        AE là cạnh chung

suy ra tam giác ABE = tam giác ACE ( 2 cạnh góc vuông )

Ta có :

\(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\)

\(P=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)

\(P=abc\left(\frac{1}{c^3}+\frac{1}{a^3}+\frac{1}{b^3}\right)\)

Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(-\frac{1}{c}\right)=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=0\)

\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(1\right)\)

Thay ( 1 ) và P ta được :

\(P=abc.\frac{3}{abc}\)

\(P=3\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc+ac+ab}{abc}=0\Leftrightarrow ab+bc+ac=0\)

Ta có

\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3}{\left(abc\right)^2}\) (1)

Ta có

\(\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3-3\left(abc\right)^2=\)

\(=\left(ab+bc+ac\right)\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2-abbc-bcac-abac\right]=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3\left(abc\right)^2\) Thay vào (1)

\(\Rightarrow\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{3\left(abc\right)^2}{\left(abc\right)^2}=3\left(đpcm\right)\)

Chọn B

Please! Help me

cái chữ quái gì vậy

(x^3-3x^2+3x-1):(x-1) uwu

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

= \(\left(x-1\right)^3:\left(x-1\right)\)

=\(\left(x-1\right)^2=\left(x^2-2x+1\right)\) ( đoạn này thích ghi ghi thì ghi khum ghi thì đừng ghi =)) )

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

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

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

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

\(=y^3-x^3\)

\(3x^2-48x=0\)

\(\Leftrightarrow3x\left(x-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=16\end{cases}}\)

Vậy \(S=\left\{0;16\right\}\).

khi x khác 0

\(\frac{2x}{x^2-16}\) xác định khi \(x\ne\pm4\)