tìm x biết
1/27x\(^2\)(x+1)-(3x+1)\(^3\)=-8
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HH
2
19 tháng 7 2021
a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)
c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)
12 tháng 1 2023
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
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 \(x=\dfrac{2}{3}\)
TD
1
19 tháng 10 2021
\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{4}\right)^2\)
\(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
\(-x^3+3x^2-3x+1=\left(-x+1\right)^3\)
NT
4
10 tháng 7 2019
\(27x^2\left(x+1\right)-\left(3x+1\right)^3=-8\)
\(\Rightarrow27x^3+27x^2-27x^3-27x^2-9x-1=-8\)
\(\Rightarrow-9x-1=-8\)
\(\Rightarrow-9x=-7\)
\(\Rightarrow x=\frac{7}{9}\)
10 tháng 7 2019
\(27x^2\left(x+1\right)-\left(3x+1\right)^3\)
\(27x^3+27^2-27x^3-27x^2-9x-1=-8\)
\(-9x-1=-8\)
\(-9x=-7\)
\(x=\frac{7}{9}\)
30 tháng 7 2020
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
30 tháng 1 2024
\(\left(3x^2-x+1\right)\left(3x^2+5x+1\right)=27x^2\)
=>\(\left(3x^2+1-x\right)\left(3x^2+1+5x\right)=27x^2\)
=>\(\left(3x^2+1\right)^2+4x\left(3x^2+1\right)-5x^2-27x^2=0\)
=>\(\left(3x^2+1\right)^2+4x\left(3x^2+1\right)-32x^2=0\)
=>\(\left(3x^2+1+8x\right)\left(3x^2+1-4x\right)=0\)
=>\(\left(3x^2+8x+1\right)\left(3x-1\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}3x^2+8x+1=0\\3x-1=0\\x-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=1\\x=\dfrac{-4\pm\sqrt{13}}{3}\end{matrix}\right.\)
MT
7 tháng 10 2019
a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
tìm x biết
\(\dfrac{1}{27x^2\left(x+1\right)-\left(3x+1\right)^3}=-8\)
Tìm dc x = \(\dfrac{-7}{72}\)
\(\dfrac{1}{27x^2\left(x+1\right)-\left(3x+1\right)^3}=-8\)
\(\Leftrightarrow\) \(-8.\left(27x^2\left(x+1\right)-\left(3x+1\right)^3\right)\) = 1
\(\Leftrightarrow\) \(-8.\left(27x^3+27x^2-27x^3-27x^2-9x-1\right)\) =1
\(\Leftrightarrow\) \(72x+8=1\)
\(\Leftrightarrow72x=\)-7
\(\Leftrightarrow x=-\dfrac{72}{7}\)
Vậy x = \(-\dfrac{72}{7}\)