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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
b.
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x³y + x - y - 1
= (x³y - y) + (x - 1)
= y(x³ - 1) + (x - 1)
= y(x - 1)(x² + x + 1) + (x - 1)
= (x - 1)[y(x² + x + 1) + 1]
= (x - 1)(x²y + xy + y + 1)
b) x²(x - 2) + 4(2 - x)
= x²(x - 2) - 4(x - 2)
= (x - 2)(x² - 4)
= (x - 2)(x - 2)(x + 2)
= (x - 2)²(x + 2)
c) x³ - x² - 20x
= x(x² - x - 20)
= x(x² + 4x - 5x - 20)
= x[(x² + 4x) - (5x + 20)]
= x[x(x + 4) - 5(x + 4)]
= x(x + 4)(x - 5)
d) (x² + 1)² - (x + 1)²
= (x² + 1 - x - 1)(x² + 1 + x + 1)
= (x² - x)(x² + x + 2)
= x(x - 1)(x² + x + 2)
e) 6x² - 7x + 2
= 6x² - 3x - 4x + 2
= (6x² - 3x) - (4x - 2)
= 3x(2x - 1) - 2(2x - 1)
= (2x - 1)(3x - 2)
f) x⁴ + 8x² + 12
= x⁴ + 2x² + 6x² + 12
= (x⁴ + 2x²) + (6x² + 12)
= x²(x² + 2) + 6(x² + 2)
= (x² + 2)(x² + 6)
g) (x³ + x + 1)(x³ + x) - 2
Đặt u = x³ + x
x³ + x + 1 = u + 1
(u + 1).u - 2
= u² + u - 2
= u² - u + 2u - 2
= (u² - u) + (2u - 2)
= u(u - 1) + 2(u - 1)
= (u - 1)(u + 2)
= (x³ + x - 1)(x³ + x + 2)
= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)
= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]
= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]
= (x³ + x - 1)(x - 1)(x² - x + 2)
h) (x + 1)(x + 2)(x + 3)(x + 4) - 1
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1
= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)
Đặt u = x² + 5x + 4
u + 2 = x² + 5x + 6
(1) u.(u + 2) - 1
= u² + 2u - 1
= u² + 2u + 1 - 2
= (u² + 2u + 1) - 2
= (u + 1)² - 2
= (u + 1 + √2)(u + 1 - √2)
= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)
= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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cái này thì "bó tay"