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Bạn ghi lộn đề rồi \(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}\) chứ không phải \(\left(\dfrac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)^{2014}\)
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2-1}\right)}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\left|\sqrt{2}-1\right|}{2}=\dfrac{\sqrt{2}-1}{2}\)
Vậy ta có \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)Ta lại có \(\left(4x^5+4x^4-x^3+1\right)^{19}=\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=\left(x^3.0+1\right)^{19}=1^{19}=1\)(1)
\(\left(\sqrt{4x^5+4x^4-5x^3+5x+3}\right)^3=\left(\sqrt{4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1+4}\right)^3=\left(\sqrt{x^3\left(4x^2+4x-1\right)-x^2\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\right)^3=\left(\sqrt{x^3.0+x^2.0+0+4}\right)^3=\left(\sqrt{4}\right)^3=2^3=8\left(2\right)\)
\(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}=\left[\dfrac{1-\sqrt{2}.\dfrac{\sqrt{2}-1}{\sqrt{2}}}{\sqrt{2.\dfrac{3-2\sqrt{2}}{4}+\sqrt{2}-1}}\right]^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}}{2}+\sqrt{2}-1}}\right)^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}+2\sqrt{2}-2}{2}}}\right)^{2014}=\left(\dfrac{\dfrac{\dfrac{1}{\sqrt{2}}}{1}}{\sqrt{2}}\right)^{2014}=1^{2014}=1\left(3\right)\)
Cộng (1),(2),(3) theo vế ta được A=1+8+1=10
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=10
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!~
b)\(x+\dfrac{4}{x+2}=3\) (ĐKXĐ: \(x\ne-2\))
\(\Leftrightarrow x+\dfrac{4}{x+2}-3=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)+4-3\left(x+2\right)}{x+2}=0\\ \Leftrightarrow\dfrac{x^2+2x+4-3x-6}{x+2}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{x+2}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x=\dfrac{-\left(-1\right)\pm\sqrt{\left(-1\right)^2-4.1.\left(-2\right)}}{2.1}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{1+8}}{2}\\ \Leftrightarrow x=\dfrac{1\pm\sqrt{9}}{2}\\ \Leftrightarrow x=\dfrac{1\pm3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+3}{2}\\x=\dfrac{1-3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy...
\(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{1}{2}\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\frac{\sqrt{2}-1}{2}\)
\(\Leftrightarrow2x+1=\sqrt{2}\)
\(\Leftrightarrow4x^2+4x-1=0\)
Giờ thế vô A đi
bạn ơi cho hỏi sao chỗ kia từ dòng số 2 sao xuống dòng số 3 được vậy
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{1}{2}.\left|\sqrt{2}-1\right|=\dfrac{\sqrt{2}-1}{2}\)
Vậy \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)
Ta lại có \(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2017}+2019=\left(4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1-1\right)^{2017}+2019=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)-1\right]^{2017}+2019=\left(x^3.0-x.0+0-1\right)^{2017}+2019=\left(-1\right)^{2017}+2019=-1+2019=2018\)
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=2018
thanks nha