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\(5x^3-45x=0\)
\(\Leftrightarrow5x\left(x^2-9\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)\left(x+3\right)=0\)
\(TH1:5x=0\)
\(\Leftrightarrow x=0\)
\(TH2:x-3=0\)
\(\Leftrightarrow x=3\)
\(TH3:x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy phương trình có nghiệm: \(S=\left\{0;\pm3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: 5x3 – 3x2 + 10x – 6 = (5x3 + 10x )+ ( -3x2– 6)
= 5x(x2 + 2) – 3(x2 + 2) = (x2 + 2)(5x – 3)
Vậy (x2 + 2)(5x – 3) = 0 ⇒ 5x – 3 = 0 (vì x2 + 2 ≥ 0, với mọi x)
⇒x = 3/5
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1) x³ + 2x² + x
= x(x² + 2x + 1)
= x(x + 1)²
2) 5x³ - 10x² + 5x
= 5x(x² - 2x + 1)
= 5x(x - 1)²
3) 8x²y - 8xy + 2x
= 2x(4xy - 4y + 1)
5) 2x² + 5x³ + x²y
= x²(2 + 5x + y)
6) 4x²y - 8xy² + 18x²y²
= 2xy(2x - 4y + 9xy)
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a.
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
f.
\(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-3\left(x-1\right)\right]\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)
5x3 - 45x = 0
=> 5x(x2- 9)=0
=>\(\orbr{\begin{cases}5x^2=0\\x^2-9=0\end{cases}}\)=>\(\orbr{\begin{cases}x^2=0\\x^2=9\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=3\\x=-3\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=3\\x=-3\end{cases}}\end{cases}}\)
5x^3- 45x=0
5x(x^2-9)=0
=> 5x =0 => x=0
hoặc x^2-9=0
=> x^2= 9 => x=-3
hoặc x=3
vậy x= 0 hoặc 3 hoặc -3