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Bài 1:
a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)
c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9,=\left(5+2x\right)^3\\ 10,=\left(y+4\right)^3\\ 11,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\\ 12,=\left(x+5y\right)\left(x^2-5xy+25y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
B) \(x^3-\dfrac{1}{8}\)
\(=x^3-\left(\dfrac{1}{2}\right)^3\)
\(=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
C) \(8x^3+y^3\)
\(=\left(2x\right)^3+y^3\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
D) \(8x^3-27y^3\)
\(=\left(2x\right)^3-\left(3y\right)^3\)
\(=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
a)\(\left(x+3\right)\left(x^2-3x+9\right)\)
b)\(\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c)\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d)\(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)8x3 + * + * + 27y3 = (* + *)3
=>A=(2x+3y)^3
b) (2x+1)^3
c)(x-2y)^3
d)(3x-2)(3x+2)
e)(3x-1)(9x^2+3x+1)
f)....................
6: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)
7: \(\left(2x+1\right)^2=4x^2+4x+1\)
8: \(\left(2x-1\right)^2=4x^2-4x+1\)
9: \(9-16x^2=\left(3-4x\right)\left(3+4x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d. 8x3 - 50x = 0
<=> 2x(4x - 25) = 0
<=> \(\left[{}\begin{matrix}2x=0\\4x-25=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{25}{4}\end{matrix}\right.\)
e. (4x - 3)2 - 3x(3 - 4x) = 0
<=> (4x - 3)2 + 3x(4x - 3) = 0
<=> (4x - 3)(4x - 3 + 3x) = 0
<=> (4x - 3)(7x - 3) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\7x-3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
d) \(8x^3-50x=0\Rightarrow2x\left(4x^2-25\right)=0\)
\(\Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\2x+5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
e) \(\left(4x-3\right)^2-3x\left(3-4x\right)=0\)
\(\Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\)
\(\Rightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\7x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a,`
`(2x - 1)^2 - 25 = 0`
`<=> (2x - 1)^2 = 25`
`<=> (2x - 1)^2 = (+-5)^2`
`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, `S = {-2; 3}`
`b,`
`8x^3 - 50x = 0`
`<=> x(8x^2 - 50) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)
Vậy, `S = {-5/2; 0; 5/2}.`
a) (2x - 1)² - 25 = 0
(2x - 1)² - 5² = 0
(2x - 1 - 5)(2x - 1 + 5) = 0
(2x - 6)(2x + 4) = 0
2x - 6 = 0 hoặc 2x + 4 = 0
*) 2x - 6 = 0
2x = 6
x = 3
*) 2x + 4 = 0
2x = -4
x = -2
Vậy x = -2; x = 3
b) 8x³ - 50x = 0
2x(4x² - 25) = 0
2x[(2x)² - 5²] = 0
2x(2x - 5)(2x + 5) = 0
2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0
*) 2x = 0
x = 0
*) 2x - 5 = 0
2x = 5
x = 5/2
*) 2x + 5 = 0
2x = -5
x = -5/2
Vậy x = -5/2; x = 0; x = 5/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(2\left(x-2\right)^3=2-x\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b: ta có: \(8x^3-72x=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
\(2^2\cdot2^2=2^{2+2}=2^4=16\)
\(\left(8\cdot3\right)^2=8^2\cdot3^2=64\cdot9=576\)