Giải PT : x² - 3x + 1 = 0. thay x vào là giải đc

\(ĐKXĐ:\)\(x\ne\left\{0;1;2;3;4;5\right\}\)

\(P=\frac{1}{x^2-x}+\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}\)

\(=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}\)

\(=\frac{1}{x-5}-\frac{1}{x}\)

\(=\frac{5}{x\left(x-5\right)}\)

Ta có:     \(x^3-x^2+2=0\)

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

Xét:    \(x^2-2x+2=\left(x-1\right)^2+1\)\(>0\)

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

\(\Leftrightarrow\)\(x=-1\)(t/m)

Vậy   tại     \(x=-1\)  thì:

          \(P=\frac{5}{-1\left(-1-5\right)}=\frac{5}{6}\)

ĐKXĐ \(x\ne0,1,2,3,4,5\)

\(P=\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(P=\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+...+\frac{1}{x-5}-\frac{1}{x-4}\)

\(P=\frac{1}{x-5}-\frac{1}{x}\)

\(P=\frac{5}{x\left(x-5\right)}\)

\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{6+3x\left(x^2-x-1\right)+2015-4x^3}\)

Theo bài ra: \(x^2-x-1=0\)

\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)

Vậy:...

Mk nhầm đoạn số 6 bạn sửa lại thành x^6 nhé:

\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}\)

Theo bài ra: \(x^2-x-1=0\)

\(\Rightarrow\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)

Vậy:......

kêu gọi giúp mà bn minh anh làm hay như z mà k đúng trả ơn hả?, tui đúng cho

Ta có: \(\frac{x}{x^2+x+1}=\frac{1}{4}\Leftrightarrow4x=x^2+x+1\Leftrightarrow x^2-3x+1=0\)

\(A=\frac{\left(x^5-3x^4+x^3\right)+\left(3x^4-9x^3+3x^2\right)+\left(5x^3-15x^2+5x\right)+\left(12x^2-36x+12\right)+21x}{\left(x^4-3x^3+x^2\right)+\left(3x^3-9x^2+3x\right)+\left(15x^2-45x+15\right)+42x}\)

\(A=\frac{21x}{42x}=\frac{1}{2}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)

\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)

à nhầm, \(a=b=c=\frac{4}{3}\) nhé