Tìm x : x^3 + 27 = -x^2 + 9
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\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)
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(x + 3)(x2 - 3x + 9) - x(x - 2)2 = 27
\(\Leftrightarrow\) x3 + 27 - x( x2 - 4x + 4) = 27
\(\Leftrightarrow\) x3 + 27 - x3 + 4x2 - 4x - 27 = 0
\(\Leftrightarrow\) 4x2 - 4x = 0
\(\Leftrightarrow\) 4x ( x - 1) = 0
khi 4x = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) x = 0 \(\Leftrightarrow\) x = 1
Chúc bạn học tốt
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.2^2=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+4x=27\\ 4x+27=27\\ 4x=0\\ x=0\)
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\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
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ĐKXĐ: \(x=\pm3\)
Nếu \(x=3\), phương trình tương đương
\(x^3+\sqrt{x^2-9}-\sqrt{9-x^2}-27=0\)
\(\Leftrightarrow0=0\)
\(\Rightarrow x=3\) là nghiệm của phương trình
Nếu \(x=-3\), phương trình tương đương
\(x^3+\sqrt{x^2-9}-\sqrt{9-x^2}-27=0\)
\(\Leftrightarrow-54=0\)
\(\Rightarrow x=-3\) không phải là nghiệm của phương trình
Vậy ...
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\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
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\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-\dfrac{2}{3}\)
=>x*1/3=-8/3
hay x=-8
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x-1\right)^2+5=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=-3\)
\(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\)
hay x=-3