(x-5)(x\(^2\)+26) +(5-x)(1-5x)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = ( x - 5 ) ( x2 + 26 + 5x - 1 )
A = ( x - 5 ) ( x2 + 5x + 25 )
A = ( x - 5 )3
\(\text{a. A = (x - 5)(x^2 + 26) + (x - 5)(5x - 1)}\\ A=\left(x-5\right)\left(x^2+5x+25\right)\\ A=x^3-5^3\\ A=x^3-125\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,x^3-3x^2=0\)
\(x^2\left(x-3\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)
\(2,3x^3-48x=0\)
\(3x\left(x^2-16\right)=0\)
\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)
\(3,5x\left(x-1\right)=x-1\)
\(5x^2-5x=x-1\)
\(5x^2-6x+1=0\)
\(5x^2-5x-x+1=0\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)
\(4,2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\)
\(-x^2-3x+10=0\)
\(-x^2-5x+2x+10=0\)
\(-x\left(x+5\right)+2\left(x+5\right)=0\)
\(\left(x+5\right)\left(2-x\right)=0\)
\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)
\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x-26=0\)
\(-13\left(x+2\right)=0\)
\(x=-2\left(TM\right)\)
Trả lời:
1, \(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
Vậy x = 0; x = 3 là nghiệm của pt.
2, \(3x^3-48x=0\)
\(\Leftrightarrow3x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)
Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.
3, \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = 1; x = 1/5 là nghiệm của pt.
4, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)
Vậy x = - 5; x = 2 là nghiệm của pt.
5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
\(\Leftrightarrow-13x=26\)
\(\Leftrightarrow x=-2\)
Vậy x = - 2 là nghiệm của pt.
f, (x+3)3
:3-1=-10
(x+3)3
:3=-10+1
(x+3)3
:3=-9
(x+3)3
:3=33
x+3:3=3