Tìm x, biết:
\(4\left(x^2+x+1\right)^3=27x^2\left(x-1\right)^2+\left(x-1\right)^2\left(2+x\right)\left(2x+1\right)^2\)
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Những câu hỏi liên quan
15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
AM
14 tháng 7 2018
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
TH
29 tháng 8 2018
a) \(27x^3+27x^2+9x+1=64\)
\(\Rightarrow27x^3+27x^2+9x-63=0\)
\(\Rightarrow27x^3-27x^2+54x^2-54x+63x-63=0\)
\(\Rightarrow27x^2\left(x-1\right)+54x\left(x-1\right)+63\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(27x^2+54x+63\right)=0\)
\(\Rightarrow\left(x-1\right).9\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x^2+6x+7=0\end{matrix}\right.\)
Mà ta có:
\(3x^2+6x+7\)
\(=3\left(x^2+2x+\dfrac{7}{3}\right)\)
\(=3\left(x^2+2x+1-1+\dfrac{7}{3}\right)\)
\(=3\left(x+1\right)^2+4\)
Vì \(3\left(x+1\right)^2\ge0\) với mọi x
\(\Rightarrow3\left(x+1\right)^2+4\ge4\)
\(\Rightarrow3x^2+6x+7\) vô nghiệm
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Rightarrow12x-8=4\)
\(\Rightarrow12x=12\)
\(\Rightarrow x=1\)
c) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x-2\right)\left(x+2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-\left(x^3+3^3\right)+3\left(x^2-2^2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-x^3-9+3x^2-12=2\)
\(\Rightarrow3x-22=2\)
\(\Rightarrow3x=24\)
\(\Rightarrow x=8\)