\(x^2-5x+1=0\)

\(\left(a=1;b=-5;c=1\right)\)

\(\Delta=b^2-4ac\)

\(\Delta=\left(-5\right)^2-4.1.1\)

\(\Delta=21>0\)

\(\sqrt{\Delta}=\sqrt{21}\)

Phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{21}}{2.1}=\frac{5+\sqrt{21}}{2}\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{21}}{2.1}=\frac{5-\sqrt{21}}{2}\)

Thay  \(x=\frac{5+\sqrt{21}}{2}v\text{ào}M=\frac{x^4+x^2+1}{2x^2}\) ta được : 

                                               \(M=\frac{\left(\frac{5+\sqrt{21}}{2}\right)^2+\left(\frac{5+\sqrt{21}}{2}\right)^2+1}{2.\left(\frac{5+\sqrt{21}}{2}\right)^2}=12\)

Thay \(x=\frac{5-\sqrt{21}}{2}v\text{ào}M=\frac{x^4+x^2+1}{2x^2}\) ta được : 

                                               \(M=\frac{\left(\frac{5-\sqrt{21}}{2}\right)^4+\left(\frac{5-\sqrt{21}}{2}\right)^2+1}{2.\left(\frac{5-\sqrt{21}}{2}\right)^2}=12\)

Vậy : M = 12 

a) A xác định \(\Leftrightarrow x^2-4\ne0\Leftrightarrow\left(x-2\right)\left(x+2\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)

b) \(A=\frac{x^2+4x+4}{x^2-4}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{x-2}\)

c) Để A nguyên thì x + 2 ⋮ x - 2

<=> x - 2 + 4 ⋮ x - 2

Vì x - 2 ⋮ x - 2

=> 4 ⋮ x - 2=> x - 2 thuộc Ư(4) = { 1; 2; 4; -1; -2; -4 }

Tự giải nốt nhé

\(a,ĐKXĐ:x\ne\pm2\)

\(b,A=\frac{x^2+4x+4}{x^2-4}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{x-2}\)

\(c,\frac{x+2}{x-2}=\frac{x-2+4}{x-2}=1+\frac{4}{x-2}\)

Do đó : A có giá trị nguyên \(\Leftrightarrow\frac{4}{x-2}\) có giá trị nguyên 

                    \(\Leftrightarrow4⋮x-2\) ( vì \(\left(x-2\right)\inℤ\) )

                     \(\Leftrightarrow\left(x-2\right)\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Bn lập bảng xét các giá trị để tìm x

\(\hept{\begin{cases}\left(a^3-3ab^2\right)^2=25\\\left(b^3-3a^2b\right)^2=100\end{cases}}\Leftrightarrow\hept{\begin{cases}a^6-6a^4b^2+9a^2b^4=25\\b^6-6a^2b^4+9a^4b^2=100\end{cases}}\)

Cộng 2 đẳng thức lại ta được:

\(a^6+3a^4b^2+3a^2b^4+b^6=125\Leftrightarrow\left(a^2+b^2\right)^3=125\Leftrightarrow a^2+b^2=5\)

\(\Rightarrow P=2018\left(a^2+b^2\right)=2018.5=...\)

Ta có : \(a^3-3ab^2=5\)

\(\Rightarrow\left(a^3-3ab^2\right)^2=a^6-6a^4b^2+9a^2b^4=25\)

Và \(b^3-3a^2b=10\)

\(\Rightarrow\left(b^3-3a^2b\right)^2=b^6-6a^4b^2+9a^4b^2=100\)

Suy ra : \(a^6++3a^2b^4+3a^4b^2+b^6=125\)

Hoặc : \(\left(a^2+b^2\right)^3=125\Rightarrow a^2+b^2=5\)

Do đó : \(P=2018a^2+2018b^2=2018\left(a^2+b^2\right)=2018.5=10090\)

ĐKXĐ: x khác 1,-1

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

\(=\frac{1+x-1-x-1}{\left(x-1\right).\left(x+1\right)}=\frac{-1}{\left(x-1\right).\left(x+1\right)}\)