Ta có 

\(MN\perp BC;AB\perp BC\) => MN//AB \(\Rightarrow\frac{MN}{AB}=\frac{CM}{CA}\) (Talet trong tam giác)

\(MP\perp AD;CD\perp AD\) => MP//CD \(\Rightarrow\frac{MP}{CD}=\frac{AM}{CA}\) (Talet trong tam giác)

\(\Rightarrow\frac{MN}{AB}+\frac{MP}{CD}=\frac{CM}{CA}+\frac{AM}{CA}=\frac{CA}{CA}=1\left(dpcm\right)\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)(vì \(a+b+c\ne0\))

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

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

\(M=\frac{x^2}{x^4+x^2+1}=\frac{x^2}{x^4+2x^2+1-x^2}=\frac{x^2}{\left(x^2+1\right)^2-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

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

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

Suy ra \(M=a.\frac{a}{2a+1}=\frac{a^2}{2a+1}\).

\(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)\left(ax+by+cz\right)=0\)

\(\Leftrightarrow ax^2+by^2+cz^2+axy+axz+bxy+byz+cxz+cyz=0\)

\(\Leftrightarrow ax^2+by^2+cz^2+xz\left(a+c\right)+xy\left(a+b\right)+yz\left(b+c\right)=0\)

\(\Leftrightarrow ax^2+by^2+cz^2-bxz-cxy-ayz=0\)

\(\Leftrightarrow ax^2+by^2+cz^2-xyz\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=0\)

\(\Leftrightarrow ax^2+by^2+cz^2=0\)

Từ \(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x^2=\left(y+z\right)^2\\y^2=\left(z+x\right)^2\\z^2=\left(x+y\right)^2\end{cases}}\)

\(\Rightarrow ax^2+by^2+cz^2\)

\(=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)

\(=ay^2+2ayz+az^2+bz^2+2bzx+bx^2+cx^2+2cxy+cy^2\)

\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\left(1\right)\)

Từ \(a+b+c=0\)

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

Thay \(\left(2\right)\)vào \(\left(1\right)\), ta được :

\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2+2\left(ayz+bzx+cxy\right)\)

Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)\(\Rightarrow ayz+bzx+cxy=0\)

\(\Rightarrow ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\)

\(\Rightarrowđpcm\)

\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right)\left(x+3\right)}=0\)(ĐK: \(x\ne-1,x\ne-3\)) 

\(\Leftrightarrow\frac{x\left(2x+2\right)\left(x+3\right)+x\left(2x-6\right)\left(x+3\right)-4x\left(2x-6\right)}{\left(2x-6\right)\left(2x+2\right)\left(x+3\right)}=0\)

\(\Rightarrow4x^3+12x=0\)

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

\(\Leftrightarrow x=0\)(thỏa).

Ta có \(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=\frac{1}{a-b-c}\)

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

=> \(\frac{b-a}{ab}=\frac{a-b}{\left(a-b-c\right)c}\)

Khi b - a = 0

=> (b - a)(a - c)(b + c) = 0 (1)

Khi b - a \(\ne0\)

=> ab = -(a - b - c).c

=> ab = -ac + bc + c² 

=> ab + ac - bc - c² = 0

=> a(b + c) - c(b + c) = 0

=> (a - c)(b + c) = 0

=> (b - a)(a - c)(b + c) = 0 (2)

Từ (1)(2) => (b - a)(a - c)(b + c) = 0

=> b - a = 0 hoặc a - c = 0 hoặc b + c = 0

=> a = b hoặc a = c hoặc b = -c

Vậy tồn tại 2 số bằng nhau hoặc đối nhau