\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\frac{100+x}{98}+\frac{100+x}{96}-\frac{100+x}{94}-\frac{100+x}{92}=0\)

\(\Rightarrow\left(100+x\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\ne0\)

\(\Rightarrow100+x=0\)

\(\Rightarrow x=-100\)

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}< \frac{1}{96}< \frac{1}{94}< \frac{1}{92}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)

Vậy \(x+100=0\Leftrightarrow x=-100\)

\(M=x^2+x+10\)

\(=x^2+x+\frac{1}{4}+\frac{39}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\)

Vậy \(M_{min}=\frac{39}{4}\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

\(M=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{39}{4}\)

\(M=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\)

\(\left(x+\frac{1}{2}\right)^2+\frac{39}{4}\ge0\)​

\(\Rightarrow M\ge\frac{39}{4}\)

Dấu "=" xảy ra: \(\left(x+\frac{1}{2}\right)^2=0\)

                            \(x+\frac{1}{2}=0\)

                            \(x=-\frac{1}{2}\)

Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)

\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)

\(\Rightarrow bcx+acy+abz=0\)

Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)

\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)

\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))

Từ (1) \(\Rightarrow bcx+acy+abz=0\)

Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)

Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)

\(=4\)

\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)

Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)

Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)

\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)

\(=-\frac{3}{2}\)

\(P\left(x\right)=4x^4+1\)

\(=4x^4+4x^2+1-4x^2\)

\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)

\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)

\(P\left(x\right)=4x^4+1\)

              \(=\left(\sqrt{4}x^2\right)^2+1^2\)

                \(=\left(2x^2\right)^2+1^2\)

               \(=\left(2x^2+1\right)^2-4x^2\)

                \(=\left(2x^2+1\right)^2-\left(2x\right)^2\)

                  \(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

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

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

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

\(\Rightarrow x=0;x=1\)

\(\left(x+1\right)^3-x^3-1=0\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x^3+1\right)=0\)

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

\(\Leftrightarrow\left(x+1\right)\cdot3x=0\)

\(\Leftrightarrow x=0;x=-1\)

\(x\left(x-2015\right)-\left(x-2015\right)=0\)

\(\left(x-1\right)\left(x-2015\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2015=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2015\end{cases}}\)

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