bài 1 tìm x
3x (x - 2) - x +2 = 0
bài 2 tìm x
x^2 - 6x^2 +12x - 8 =0
16x^ - 9 (x +1)^2 =0
27+27x - 9x^2 +x^3=0
cac cậu giup mk voi mk cần gâp cho ngày mai
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HT
2
13 tháng 1 2022
2: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3 hoặc x=2
5: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{-2;1;-1\right\}\)
3 tháng 11 2019
1a) 5x(x - 3) - x + 3 = 0
=> 5x(x - 3) - (x - 3) = 0
=> (5x - 1)(x - 3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}5x=1\\x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
b) x3 + 3x2 = -3x - 1
=> x3 + 3x2 + 3x + 1 = 0
=> (x + 1)3 = 0
=> x + 1 = 0
=> x = -1
DT
0
7 tháng 9 2016
a ) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
b ) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)=0\)
\(\Leftrightarrow x=3\)
7 tháng 9 2016
a) x3 - 6x2 + 12x - 8 = 0
( x - 2 ) 3 = 0
x - 2 = 0
x = 2
b) x3 + 9x2 + 27x + 27 = 0
( x + 3 )3 = 0
x + 3 = 0
x = -3
4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
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: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
6 tháng 9 2016
a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)
b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)
29 tháng 7 2017
c. \(8x^3-12x^2+6x-1=0\)
\(\Rightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow x=\frac{1}{2}\)
16 tháng 10 2020
Bài 1: Tìm x
a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=0\)
\(\Leftrightarrow-12x-24=0\)
\(\Leftrightarrow-12x=24\)
hay x=-2
Vậy: x=-2
b) Ta có: \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)
\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)-11\left(x-1\right)\left(x+1\right)-6=0\)
\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11\left(x^2-1\right)-6=0\)
\(\Leftrightarrow11x^2+6x+12-11x^2+11=0\)
\(\Leftrightarrow6x+23=0\)
\(\Leftrightarrow6x=-23\)
hay \(x=-\frac{23}{6}\)
Vậy: \(x=-\frac{23}{6}\)
c) Ta có: \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
hay \(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
d) Ta có: \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Vậy: x=-3
KV
16 tháng 10 2020
a) (2x + 1)2 - 4(x + 2)2 = 9
4x2 + 4x + 1 - 4(x2 + 4x + 4) = 9
4x2 + 4x + 1 - 4x2 - 16x - 16 = 9
-12x - 15 = 9
-12x = 9 + 15
-12x = 24
x = 12 : (-2)
x = -2
b) (3x - 1)2 + 2(x + 3)2 + 11(x + 1)(1 - x) = 6
9x2 - 6x + 1 + 2(x2 + 6x + 9) - 11(x + 1)(x - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11(x2 - 1) = 6
9x2 - 6x + 1 + 2x2 + 12x + 18 - 11x2 + 11 = 6
6x + 30 = 6
6x = 6 - 30
6x = -24
x = -24 : 6
x = -4
c) 8x3 - 12x2 + 6x - 1 = 0
(2x)3 - 3.(2x)2.1 + 3.2x.12 - 13 = 0
(2x - 1)3 = 0
2x - 1 = 0
2x = 1
x = 1/2
d) x3 + 9x2 + 27x + 27 = 0
x3 + 3.x2.3 + 3.x.32 + 33 = 0
(x + 3)3 = 0
x + 3 = 0
x = 0 - 3
x = -3
\(3x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
\(B1:\)
\(3x\left(x-2\right)-\left(x-2\right)=0\)
\(\left(3x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)