Ta có : \(Q=\frac{x^6-3x^5+3x^4-x^3+2020}{x^6-x^3-3x^2-3x+2020}\)

=> \(Q=\frac{\left(x^6-x^5-x^4\right)+\left(-2x^5+2x^4+2x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(-x^3+x^2+x\right)+\left(x^2-x-1\right)+2021}{\left(x^6-x^5-x^4\right)+\left(x^5-x^4-x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(2x^3-2x^2-2x\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4\left(x^2-x-1\right)-2x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}{x^4\left(x^2-x-1\right)+x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}\)

=> \(Q=\frac{x^4.0-2x^3.0+2x^2.0-x.0+0+2021}{x^4.0+x^3.0+2x^2.0+0+2021}\)

=> \(Q=\frac{2021}{2021}=1\)

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

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

Theo đề ta có

\(x=2-\sqrt{3}\)

\(\Rightarrow\left(4-x\right)x=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1\)

Q = x⁵ - 3x⁴ - 3x³ + 6x² - 20x + 2020

= (x⁵ - 4x⁴) + (x⁴ - 4x³) + (x³ - 4x²) + (10x² - 40x) + 20x + 2020

= - x³ - x² - x - 10 + 20x + 2020

= (- x³ + 4x²) + ( - 5x² + 20x) - x + 2010

= x + 5 - x + 2010 = 2015

cau tra loi la chinh no

băng 2015

Câu 1 thay 3x =4y vào tính

Câu 2 bằng trừ 3

Chọn A

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

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

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

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

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

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

c: \(x^3-x^2+2=0\)

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

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

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

=>x+1=0

=>x=-1

Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)

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