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Ta có: B = \(\left(4x^5+4x^4-5x^3+2x-2\right)^2+2017\)
Đặt D = \(4x^5+4x^4-5x^3+2x=x\left(4x^4+4x^3-5x^2+2\right)\)
Thay \(x=\dfrac{\sqrt{5}-1}{2}\) vào D ta được:
D =
\(\dfrac{\sqrt{5}-1}{2}.\left[4\left(\dfrac{\sqrt{5}-1}{2}\right)^4+4\left(\dfrac{\sqrt{5}-1}{2}\right)^3-5\left(\dfrac{\sqrt{5}-1}{2}\right)^2+2\right]\)
D=\(\dfrac{\sqrt{5}-1}{2}\left[\dfrac{4\left(\sqrt{5}-1\right)^4}{16}+\dfrac{4\left(\sqrt{5}-1\right)^3}{8}-\dfrac{5\left(\sqrt{5}-1\right)^2}{4}+2\right]\)
D = \(\dfrac{\sqrt{5}-1}{2}\left[\dfrac{\left(\sqrt{5}-1\right)^4}{4}+\dfrac{2\left(\sqrt{5}-1\right)^3}{4}-\dfrac{5\left(\sqrt{5}-1\right)^2}{4}+\dfrac{8}{4}\right]\)
D =
\(\dfrac{\sqrt{5}-1}{2}.\left(\dfrac{25-20\sqrt{5}+30-4\sqrt{5}+1+10\sqrt{5}-30+6\sqrt{5}-2-25+10\sqrt{5}-5+8}{4}\right)\)
D = \(\dfrac{\sqrt{5}-1}{2}\left(\dfrac{2\left(\sqrt{5}+1\right)}{4}\right)\) = \(\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{4}\) = \(1\)
=> B = \(\left(1-2\right)^2+2017\) = 2018
P/s: Haha!Đây là phương an toàn mà dễ lm nhất!~
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 \(B=2022\)
Ta có:
x = \(\frac{1}{2}\)\(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
= \(\frac{1}{2}\)\(\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{1}}\)
= \(\frac{1}{2}\)(\(\sqrt{2}\)-1)
=> 2x = \(\sqrt{2}\)-1
=> (2x)2= ( \(\sqrt{2}\)-1)2
=> 4x2= 2-2\(\sqrt{2}\)+1
=> 4x2= -2( \(\sqrt{2}\)-1)+1
=> 4x2= -4x +1 => 4x2+4x-1=0
Lại có:
A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19
= [ x3( 4x2+4x-1) +1]19
=1
A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3
= (\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\))3
= 23=8
A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)
= \(\sqrt{2}\)- \(\sqrt{2}\)\(\sqrt{1-\sqrt{2}}\)
Cộng 3 số vào ta được A
\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)
\(\Leftrightarrow4x^2+4x-1=0\)
\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)
\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)
\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
\(M=1+8+1=10\)