\(\frac{x+m}{n+p}+\frac{x+n}{p+m}+\frac{x+p}{m+n}+3=0\)

\(\Leftrightarrow\frac{x+m}{n+p}+1+\frac{x+n}{p+m}+1+\frac{x+p}{m+n}+1=0\)

\(\Leftrightarrow\frac{x+m+n+p}{n+p}+\frac{x+m+n+p}{p+m}+\frac{x+m+n+p}{m+n}=0\)

\(\Leftrightarrow\left(x+m+n+p\right)\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{m+n}\right)=0\)

Dễ thấy \(\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{m+n}\right)\ne0\)

Nên x + m + n + p = 0\(\Rightarrow x=-\left(m+n+p\right)\)

\(\frac{x-12}{21}+\frac{x-10}{23}=\frac{x-8}{25}+\frac{x-6}{27}\)

\(\Leftrightarrow\frac{x-12-21}{21}+\frac{x-10-23}{23}-\frac{x-8-25}{25}-\frac{x-6-27}{27}=0\)

\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}-\frac{x-33}{25}-\frac{x-33}{27}=0\)

\(\Leftrightarrow\left(x-33\right)\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)=0\)

Vif \(\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)\ne0\)

\(\Rightarrow x-33=0\)

\(\Rightarrow x=33\)

\(\frac{x-12}{21}+\frac{x-10}{23}=\frac{x-8}{25}+\frac{x-6}{27}\)

\(\Leftrightarrow\frac{x-12}{21}+1+\frac{x-10}{23}+1=\frac{x-8}{25}+1+\frac{x-6}{27}+1\)

\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}=\frac{x-33}{25}+\frac{x-33}{27}\)

\(\Leftrightarrow\frac{x-33}{21}+\frac{x-33}{23}-\frac{x-33}{25}-\frac{x-33}{27}=0\)

\(\Leftrightarrow\left(x-33\right)\left(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\right)=0\)

Mà \(\frac{1}{21}+\frac{1}{23}-\frac{1}{25}-\frac{1}{27}\ne0\)

\(\Rightarrow x-33=0\)

\(\Leftrightarrow x=33\)

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

     \(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ca+x^2\)

     \(=4x^2-2ax-2bc-2cx+ab+bc+ca\)     

   \(=4x^2-2\left(a+b+c\right)x+ab+bc+ca\)

với \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c\Rightarrow2x=a+b+c\)

\(\Rightarrow M=\left(a+b+c\right)^2-\left(a+b+c\right)^2+ab+bc+ca\)

            \(=ab+bc+ca\)

\(x^2+\frac{81x^2}{\left(x+9\right)^2}=40^{^{\left(1\right)}}\)

\(ĐK:x\ne-9\)

\(\left(1\right)\Leftrightarrow x^2-2.x.\frac{9x}{x+9}+\frac{81x^2}{\left(x+9\right)^2}+\frac{18x^2}{x+9}=40\)

\(\Leftrightarrow\left(x-\frac{9x}{x+9}\right)^2+\frac{18x^2}{x+9}=40\)

\(\Leftrightarrow\left(\frac{x^2}{x+9}\right)^2+18.\frac{x^2}{x+9}=0\)

Đặt \(\frac{x^2}{x+9}=t\)ta có:

\(t^2-18t-40=0\)

\(\Leftrightarrow\left(t+2\right)\left(t-20\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+2=0\\t-20=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}t=-2\\t=20\end{cases}}\)

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rồi tự thay vào nha