(3x-1)3=27
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HD
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
1/
\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)
\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)
\(\Leftrightarrow43x=0\Rightarrow x=0\)
2/
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)
\(\Leftrightarrow x^3+27-x^3+x=27\)
\(\Leftrightarrow x=0\)
HT
17 tháng 4 2022
-1 5/27-(3x-7/9)3=-24/27
(3x-7/9)3 =-32/27-(-24/27)
(3x-7/9)3=-8/27
(3x-7/9)3=(-2/3)3
⇒3x-7/9=-2/3
3x =-2/3+7/9
3x =1/9
x =1/9:3
x =1/27
Chúc bạn học tốt!
DT
26 tháng 6 2016
\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)
=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)
=\(\frac{x}{x-3}\)
24 tháng 5 2022
b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
DV
10 tháng 12 2016
= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]
\(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)
= \(\frac{\left(x-3\right).\left(x+3\right)}{\left(x-3\right)^2}\)
= \(\frac{x+3}{x-3}\)
k mik nhé. Plssss~
TA
4 tháng 8 2021
`=(3x-1)^2+2(2x-1)(2x+5)+(2x+5)^2-(2x-3)(4x^2+6x+9):(2x-3)`
`=(3x-1)^2+2(2x-1)(2x+5)+(2x+5)^2-(2x-3)(4x^2+6x+9):(2x-3)`
`=(3x-1+2x+5)^2-(4x^2+6x+9)`
`=(5x+4)^2-(4x^2+6x+9)`
`=25x^2+40x+16-4x^2-6x-9`
`=21x^2+34x+7`
4 tháng 8 2021
Ta có: \(\left(3x-1\right)^2-2\left(1-3x\right)\left(2x+5\right)+\left(5+2x\right)^2-\left(8x^3-27\right):\left(2x-3\right)\)
\(=\left(3x-1+2x+5\right)^2-\left(4x^2+6x+9\right)\)
\(=\left(5x+4\right)^2-\left(4x^2+6x+9\right)\)
\(=25x^2+40x+16-4x^2-6x-9\)
\(=21x^2+34x+7\)
9 tháng 10 2020
phần b nhé
(X+0,7)3=-27
(X+0,7)3=(-3)3
X+0,7=-3
X=3-0,7=2,3
Vậy...
NQ
1 tháng 6 2021
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+
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33543537656453664
LB
7 tháng 3 2018
Đặt \(t=3x+y\)
pt \(\Leftrightarrow t^3-3t^2+3t-1=-27\)
\(\Leftrightarrow\left(t-1\right)^3+3^3=0\)
\(\Leftrightarrow\left(t-1+3\right)\left(\left(t-1\right)^2-3\left(t-1\right)+9\right)=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-5t+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t+2=0\\t^2-5t+13=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-2\\t^2-5t+13=0\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow3x+y=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2-3x\\x\in R\end{matrix}\right.\)
( 3x - 1 )3 = 27
( 3x - 1 )3 = 33
3x - 1 = 3
3x = 3 + 1
3x = 4
x = 4 : 3