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Ta có:
\(x=\frac{1}{2}.\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{\sqrt{2}-1}{2}\)
\(\Rightarrow x\left(x+1\right)=\frac{\sqrt{2}-1}{2}.\frac{\sqrt{2}+1}{2}=\frac{1}{4}\)
Thế vô bài toán ta được
\(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2016}+2017\)
\(=\left(4x^4\left(x+1\right)-5x^3+5x-2\right)^{2016}+2017\)
\(=\left(-4x^3+5x-2\right)^{2016}+2017\)
\(=\left(\left(-4x^3-4x^2\right)+\left(4x^2+4x\right)+x-2\right)^{2016}+2017\)
\(=\left(-x+1+x-2\right)^{2016}+2017\)
\(=\left(-1\right)^{2016}+2017=2018\)
\(x=\dfrac{1}{2}\cdot\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\sqrt{2}-1}{2}\)
\(A=\left[4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^4+4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^3-5\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^2+5\cdot\dfrac{\sqrt{2}-1}{2}-2\right]^{2015}+2016\)
=-1,13+2016=2014,87
\(x=\frac{\sqrt{5}-1}{2}\Leftrightarrow2x+1=\sqrt{5}\)
\(\Rightarrow4x^2+4x+1=5\)
\(\Rightarrow4x^2+4x-4=0\)
\(\Rightarrow x^2+x-1=0\)
\(\Rightarrow-x^2=x-1\Rightarrow-x^3=x^2-x\)
\(B=\left[4x^3\left(x^2+x-1\right)-x^3+2x-2\right]^2+2021\)
\(=\left(-x^3+2x-2\right)^2+2021\)
\(=\left(x^2-x+2x-2\right)^2+2021\)
\(=\left(x^2+x-1-1\right)^2+2021\)
\(=\left(-1\right)^2+2021=2022\)
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
Ta có : \(x=\frac{1}{2}\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{1}{2}\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\frac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\frac{\sqrt{2}-1}{2}\)
Thay \(x=\frac{\sqrt{2}-1}{2}\)vào \(4x^5+4x^4-5x^3+5x-2\)được kết quả bằng -1
\(\Rightarrow A=\left(-1\right)^{2012}+2103=1+2103=2104\)