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NN
15 tháng 11 2017
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
NN
15 tháng 11 2017
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
NV
8 tháng 1 2018
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
15 tháng 12 2020
Tương tự mấy phần kia
\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)
\(=\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\frac{-3+x}{\left(x-2\right)\left(x-3\right)}=\frac{-1}{x-2}\)
PN
13 tháng 6 2019
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy pt có nghiệm duy nhất x = 0.
b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)
c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)
\(\Leftrightarrow x^2-11x=0\)
\(\Leftrightarrow x\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)
d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)
\(\Leftrightarrow41-10x=1\)
\(\Leftrightarrow-10x=40\)
\(\Leftrightarrow x=-4\)
Vậy pt có nghiệm duy nhất x = -4.
e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)
\(\Leftrightarrow8x=-13\)
\(\Leftrightarrow x=-\frac{13}{8}\)
Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)
3 tháng 7 2016
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
7 tháng 12 2015
a) 3x^3-12x=0
3x(x^2-4)=0
3x(x-2)(x+2)=0
suy ra 3x=0 suy ra x=0
x-2=0 x=2
x+2=0 x= -2
b) (x-3)^2-(x-3)(3-x)^2=0
(x-3)^2-(x-3)(x-3)^2=0
(x-3)^2(1-x+3)=0
(x-3)^2(4-x)=0
suy ra x-3=0 suy ra x=3
4-x=0 x=4
a) và b) đã nhé bạn
a, \(\left(x+2\right)^3-\left(x+1\right)^3=3x^2+2\)
\(\Leftrightarrow\left(x+2-x-1\right)\left[\left(x+2\right)^2+\left(x+2\right)\left(x+1\right)+\left(x+1\right)^2\right]=3x^2+2\)
\(\Leftrightarrow x^2+4x+4+x^2+3x+2+x^2+2x+1=3x^2+2\)
\(\Leftrightarrow3x^2+9x+7=3x^2+2\Leftrightarrow9x=-5\Leftrightarrow x=-\frac{5}{9}\)
b, \(\left(x+1\right)^3+\left(2x+1\right)^3=\left(3x+2\right)^3\)
\(\Leftrightarrow\left(x+1+2x+1\right)\left[\left(x+1\right)^2-\left(2x+1\right)\left(x+1\right)+\left(2x+1\right)^2\right]=\left(3x+2\right)^3\)
\(\Leftrightarrow\left(3x+2\right)\left(x^2+2x+1-2x^2-3x-1+4x^2+4x+1\right)=\left(3x+2\right)^3\)
\(\Leftrightarrow\left(3x+2\right)\left(3x^2+3x+1\right)=\left(3x+2\right)^3\)
\(\Leftrightarrow\left(3x+2\right)\left[\left(3x+2\right)^2-3x^2-3x-1\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(9x^2+12x+4-3x^2-3x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(6x^2+9x+3\right)=0\Leftrightarrow\left(3x+2\right)\left(2x^2+3x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(2x+1\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=-\frac{1}{2};x=-1\)
Trả lời:
a) ( x + 2 )3- ( x + 1 )3 = 3x2 + 2
<=> x3 + 6x2 + 12x + 8 - ( x3 + 3x2 + 3x + 1 ) = 3x2 + 2
<=> x3 + 6x2 + 12x + 8 - x3 - 3x2 - 3x - 1 = 3x2 + 2
<=> 3x2 + 9x + 7 = 3x2 + 2
<=> 3x2 + 9x + 7 - 3x2 - 2 = 0
<=> 9x + 5 = 0
<=> x = - 5/9
Vậy x = - 5/9 là nghiệm của pt.
b, ( x + 1 )3 + ( 2x + 1 )3 = ( 3x + 2 )3
<=> x3 + 3x2 + 3x + 1 + 8x3 + 12x2 + 6x + 1 = 27x3 + 54x2 + 36x + 8
<=> 9x3 + 15x2 + 9x + 2 = 27x3 + 54x2 + 36x + 8
<=> 9x3 + 15x2 + 9x + 2 - 27x3 - 54x2 - 36x - 8 = 0
<=> - 18x3 - 39x2 - 27x - 6 = 0
<=> x = - 1; x = - 2/3; x = - 1/2
Vậy x = - 1; x = - 2/3; x = - 1/2 là nghiệm của pt.