Tìm x biết:
x^2(x-3)+3x=1
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\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)
<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)
<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)
<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)
Vì \(x^2+x+5>0\)=>x-4=0
<=>x=4
x³ - x² - x = 1/3
<=> x³ = x² + x + 1/3
<=> 3x³ = 3(x² + x + 1/3)
<=> 3x³ = 3x² + 3x + 1
<=> 3x³ + x³ = x³ + 3x² + 3x + 1
<=> 4x³ = (x + 1)³
<=> ³√(4x³) = ³√(x + 1)³
<=> ³√4.x = x + 1
<=> ³√4.x - x = 1
<=> x(³√4 - 1) = 1
<=> x = 1/(³√4 - 1)
Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow x^2-x^2+2x=-4+3\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
\(x^2\left(x+1\right)+\left(x+1\right)=y^3\)
\(\left(x+1\right)\left(x^2+1\right)=y^3\)
\(\left(x+1\right)\left(x^2+1\right)-y^3=0\)
\(\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x^2=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\kothoaman\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=-1\\y^3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=0\end{cases}}\)
Vậy x = -1, y =0
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)
\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
k cho minh
\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)
\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)
\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)
Tính ra nhé !
Ta có: x2 (x - 3) + 3x = 1
=> x3 - 3x2 + 3x - 1 = 0
=> (x - 1)3 = 0
=> x - 1 = 0
=> x = 1
\(x^2\left(x-3\right)+3x=1\)
\(\Rightarrow x^3-3x^2+3x-1=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)