Cho x = 1 + \(\sqrt[3]{2}+\sqrt[3]{4}\)
Tính M = \(x^5\)\(-4x^4+x^3-x^2-2x+2015\)
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TT
1
1 tháng 8 2020
Ta có \(x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
<=> \(\left(x-1\right)^3=6+3.\sqrt[3]{2.4}.\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
<=>\(x^3-3x^2+3x-1=6+6.\left(x-1\right)\)
<=>\(x^3-3x^2-3x-1=0\)
=> \(P=x^2\left(x^3-3x^2-3x-1\right)-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x-1+2016\)
=> \(P=2016\)
I
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 9 2021
\(x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Rightarrow\left(x-1\right)^3=6+3\sqrt[3]{8}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)\)
\(\Rightarrow x^3-3x^2+3x-1=6+6\left(x-1\right)\)
\(\Rightarrow x^3-3x^2-3x-1=0\)
\(A=x^2\left(x^3-3x^2-3x-1\right)-x^4+4x^3-2x+2019\)
\(=-x\left(x^3-3x^2-3x-1\right)+x^3-3x^2-3x+2019\)
\(=1+2019=2020\)
8 tháng 10 2016
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
5 tháng 1 2021
1) Ta có: \(\left|x^2-4x-5\right|=x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x-1\left(\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\right)\\-x^2+4x+5=x-1\left(-1< x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5-x+1=0\\-x^2+4x+5-x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=0\\-x^2+3x+6=0\end{matrix}\right.\Leftrightarrow x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{41}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{41}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{41}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}+5}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}+5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{41}+5}{2}\right\}\)
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(x=1+\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow x-1=\sqrt[3]{2}+\sqrt[3]{4}\)
\(\Rightarrow\left(x-1\right)^3=2+4+3\sqrt[3]{2.4}\left(\sqrt[3]{2}+\sqrt[3]{4}\right)=6+6\left(x-1\right)=6x\)
\(\Rightarrow x^3-3x^2+3x-1=6x\Rightarrow x^3-3x^2-3x-1=0\)
Ta có:
\(M=\left(x^5-3x^4-3x^3-x^2\right)-x^4+4x^3-2x+2015\)
\(\Rightarrow M=x^2\left(x^3-3x^2-3x-1\right)-x^4+3x^3+3x^2+x+x^3-3x^2-3x-1+2016\)
\(\Rightarrow M=-x\left(x^3-3x^2-3x-1\right)+\left(x^3-3x^2-3x-1\right)+2016\)
\(\Rightarrow M=2016\)