Ta có: \(x=2-\sqrt{3}\)\(\Rightarrow2-x=\sqrt{3}\)\(\Rightarrow\left(2-x\right)^2=3\)\(\Rightarrow4-4x+x^2=3\)\(\Rightarrow x^2-4x+1=0\)

Lại có: \(B=x^5-3x^4-3x^3+6x^2-20x+2018\)

\(\Rightarrow B=x^5-4x^4+x^4+x^3-4x^3+5x^2+x^2+20x+5+2013\)

\(\Rightarrow B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2013\)

\(\Rightarrow B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2013\)

\(\Rightarrow B=x^3\cdot0+x^2\cdot0+5\cdot0+2013=2013\)

b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)

\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)

Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)

\frac{23}{\sqrt{31+12\sqrt{3}}} x = \frac{23}{\sqrt{(3\sqrt{3}+2)^2}} x = \frac{23}{3\sqrt{3}+2} x = \frac{23(3\sqrt{3}-2)}{(3\sqrt{3}+2)(3\sqrt{3}-2)} x = \frac{23(3\sqrt{3}-2)}{27-4} x = \frac{23(3\sqrt{3}-2)}{23} x = 3\sqrt{3}-2 (x+2)^2 = (3\sqrt{3})^2 x^2+4x+4 = 27 x^2+4x-23 = 0 P = \frac{x^4+5x^3-20x^2-27x+30}{x^2+4x-21} x^2+4x-21 = x^2+4x-23+2 = 0+2 = 2 x^4+5x^3-20x^2-27x+30 = x^2(x^2+4x-23) + x(x^2+4x-23) - (x^2+4x-23) + 7 = x^2(0) + x(0) - (0) + 7 = 7 P = \frac{7}{2}

\($$ x = \frac{23}{\sqrt{31+12\sqrt{3}}} \\ x = \frac{23}{\sqrt{(3\sqrt{3}+2)^2}} \\ x = \frac{23}{3\sqrt{3}+2} \\ x = \frac{23(3\sqrt{3}-2)}{(3\sqrt{3}+2)(3\sqrt{3}-2)} \\ x = \frac{23(3\sqrt{3}-2)}{27-4} \\ x = \frac{23(3\sqrt{3}-2)}{23} \\ x = 3\sqrt{3}-2 \\ \\ (x+2)^2 = (3\sqrt{3})^2 \\ x^2+4x+4 = 27 \\ x^2+4x-23 = 0 \\ \\ P = \frac{x^4+5x^3-20x^2-27x+30}{x^2+4x-21} \\ \\ x^2+4x-21 = (x^2+4x-23)+2 = 0+2 = 2 \\ \\ x^4+5x^3-20x^2-27x+30 = x^2(x^2+4x-23) + x(x^2+4x-23) - (x^2+4x-23) + 7 \\ = x^2(0) + x(0) - (0) + 7 \\ = 7 \\ \\ P = \frac{7}{2} $$ \)

\(x^2+y^2-20x-2y+101=0\)

\(\Leftrightarrow\left(x-10\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=10\\y=1\end{cases}}\)

Thế vô là ra

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

mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia

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