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\(a)x^3-0,25=0\\ \Leftrightarrow x^3-\dfrac{1}{4}=0\\ \Leftrightarrow x^3=\dfrac{1}{4}\\ x=\dfrac{\sqrt[3]{2}}{2}\)
Vậy...
\(b)x^2-10=-25\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)
Vậy...
\(c)x^2-2x-3=0\\ \Leftrightarrow x^2+x-3x-3=0\\ \Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy...
\(d)2x^2+5x-3=0\\ \Leftrightarrow2x^2+6x-x-3=0\\ \Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy...
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a) \(0,25x^3+x^2+x=0\)
\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)
\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)
\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)
Vậy....
b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)
\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)
\(\Rightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x3-0,25x=0
x(x2-0,25)=0
x(x-0,5)(x+0,5)=0
x=0 hoặc x=0,5 hoặc x=-0,5
x3 - 0,25 x = 0
<=> x(x2 - 0,25) = 0
<=> x(x - 0,5)(x + 0,5) = 0
<=> x = 0
x - 0,5 = 0
x + 0,5 = 0
<=> x = 0
x = 0,5
x = -0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\Leftrightarrow x^2=0,25\Leftrightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\end{matrix}\right.\)
Vậy pt có 3 nghiệm là.....
Ta có:
\(x^3-0,25x=0\)
\(\Rightarrow x.\left(x^2-0,25\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
Vậy \(x\in\left\{0;0,5;-0,5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d. Áp dụng BĐT Caushy Schwartz ta có:
\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-0,25x=0\)
\(\Rightarrow x\left(x^2-0,25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
Chúc bạn học tốt!!!
\(x^2-x+0,25=0\)
\(\Rightarrow x^2-2.x.\frac{1}{2}+\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)