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

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

Pham Van Hung mình ko hiểu tại sao \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Biểu thức đâu bạn ? :)))

Sau khi ib với Đinh Lan Anh  thì \(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)

\(a,ĐKXĐ:\hept{\begin{cases}a+1\ne0\\a-1\ne0\end{cases}\Leftrightarrow a\ne\pm1}\)

\(b,P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)

       \(=\frac{2a^2+a\left(a-1\right)-a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)

       \(=\frac{2a^2+a^2-a-a^2-q}{\left(a-1\right)\left(a+1\right)}\)

       \(=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}\)

       \(=\frac{2a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}\)

      \(=\frac{2a}{a+1}\)

\(c,P=\frac{2a}{a+1}=\frac{2a+2}{a+1}-\frac{2}{a+1}=2-\frac{2}{a+1}\)

Để \(P\inℤ\)thì \(2-\frac{2}{a+1}\inℤ\)

                    \(\Leftrightarrow\frac{2}{a+1}\inℤ\)

Mà \(a\inℤ\Rightarrow a+1\inℤ\)

Ta có bảng

Kết hợp ĐKXĐ \(a\ne\pm1\)ta  được \(a\in\left\{-3;-2;0\right\}\)

Vậy //////

\(A=\frac{8x^2-24x+32}{8\left(x-1\right)^2}=\frac{x^2-10x+25+7\left(x-1\right)^2}{8\left(x-1\right)^2}=\frac{\left(x-5\right)^2}{8\left(x-1\right)^2}+\frac{7}{8}\ge\frac{7}{8}\forall x\)

Dấu "=" xảy ra khi \(x-5=0\Rightarrow x=5\)

Vậy GTNN của A là \(\frac{7}{8}\) khi x = 5

la 4 nha ban

\(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

\(\left(2x-3\right)^2-4x^2=-15\)

\(\left(2x-3\right)^2-\left(2x\right)^2=-15\)

\(\left(2x-3-2x\right)\left(2x-3+2x\right)=-15\)

\(-3.\left(4x-3\right)=-15\)

\(\Leftrightarrow4x-3=5\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

a)\(A=\frac{x^2+2x+1}{x^2+1}\)

Ta có: \(x^2+1>0\forall x\)

\(\Rightarrow A\)xác định với mọi x

\(a,ĐKXĐ\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow x\ne\pm3}\)

Ta có: \(M=\frac{3}{x-3}-\frac{6x}{9-x^2}+\frac{x}{x+3}\)

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

           \(=\frac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

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

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

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

              \(=\frac{x+3}{x-3}\)

\(b,x=\frac{1}{2}\Rightarrow M=\frac{\frac{1}{2}+3}{\frac{1}{2}-3}=-\frac{7}{5}\)