loanwngf ngoằng quá

mk biết

Ta có : \(x=\frac{\sqrt{5}-1}{2}\Rightarrow2x=\sqrt{5}-1\)

\(\Leftrightarrow2x+1=\sqrt{5}\)

\(\Rightarrow\left(2x+1\right)^2=5\)

\(\Rightarrow4x^2+4x+1=5\Rightarrow x^2+x-1=0\)

Khi đó ta có :

\(B=\left(4x^5+4x^4-5x^3+2x-2\right)^2+2021\)

\(=\left[\left(4x^5+4x^4-4x^3\right)-\left(x^3+x^2-x\right)+\left(x^2+x-1\right)-1\right]^2+2021\)

\(=\left[4x^3\left(x^2-x+1\right)-x\left(x^2+x-1\right)+\left(x^2+x-1\right)-1\right]^2+2021\)

\(=\left(-1\right)^2+2021=2022\)

Vậy \(B=2022\)

Tao Không biết làm

Mài cũng có não mà done

Bất lực, tìm được mỗi max P T.T

Đề bài là GTNN :))

= 0 bạn nhé

\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{4-\left(\sqrt{2+\sqrt{3}}\right)^2}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{2^2-\left(\sqrt{3}\right)^2}=\sqrt{4-3}=1\)

Vì \(9>5\)\(\Rightarrow\sqrt{9}>\sqrt{5}\)\(\Rightarrow3>\sqrt{5}\)\(\Rightarrow3-\sqrt{5}>0\)

mà \(3+\sqrt{5}>0\)

\(\Rightarrow\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2.\left(3+\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)^2.\left(3-\sqrt{5}\right)}\)

\(=\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=\sqrt{\left(9-5\right)\left(3-\sqrt{5}\right)}+\sqrt{\left(9-5\right).\left(3+\sqrt{5}\right)}\)

\(=\sqrt{4.\left(3-\sqrt{5}\right)}+\sqrt{4.\left(3+\sqrt{5}\right)}\)

\(=2.\sqrt{3-\sqrt{5}}+2.\sqrt{3+\sqrt{5}}\)

1. Ta có: \(9< 10\)\(\Rightarrow\sqrt{9}< \sqrt{10}\)\(\Rightarrow3< \sqrt{10}\)\(\Rightarrow3-\sqrt{10}< 0\)(1)

Vì \(3< \sqrt{10}\)\(\Rightarrow2.3< 2\sqrt{10}\)\(\Rightarrow6< 2\sqrt{10}\)\(\Rightarrow2\sqrt{10}-6>0\)(2)

Từ (1) và (2) \(\Rightarrow\sqrt{\left(3-\sqrt{10}\right)^2}+\sqrt{\left(2\sqrt{10}-6\right)^2}\)

\(=\left|3-\sqrt{10}\right|+\left|2\sqrt{10}-6\right|\)

\(=\sqrt{10}-3+2\sqrt{10}-6=3\sqrt{10}-9\)

4. Vì \(x>0\)\(\Rightarrow x.\sqrt{\frac{9}{x}}+5\sqrt{x}=\sqrt{x^2.\frac{9}{x}}+5\sqrt{x}=\sqrt{9x}+5\sqrt{x}\)

\(=3\sqrt{x}+5\sqrt{x}=8\sqrt{x}\)

Thiếu 1 phương trình :

\(4x^2-4\left(2n+1\right)x+4n^2+96mnp+1=0\)

Ta có: \(\sin18^0\approx0,3090169944\)

           \(\frac{\sqrt{5}-1}{4}\approx0,3090169944\)

 \(\Rightarrow\)\(\sin18^0=\frac{\sqrt{5}-1}{4}\)