2x-49=5*3 mũ 2 tìm
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a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
\(1,3x\left(x-7\right)+2x-14=0\)
\(\Leftrightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{2}{3}\end{cases}}}\)
Vậy x = 7; x = - 2/3 là nghiệm của pt.
\(2,x^3+3x^2-\left(x+3\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}}\)
Vậy x = - 3; x = 1; x = - 1 là nghiệm của pt.
\(3,15x-5+6x^2-2x=0\)
\(\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{5}{2}\end{cases}}}\)
Vậy x = 1/3; x = - 5/2 là nghiệm của pt.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^3:\left|x-2\right|=2\)
\(\Leftrightarrow8:\left|x-2\right|=2\)
\(\Leftrightarrow\left|x-2\right|=8:2\)
\(\Leftrightarrow\left|x-2\right|=4\)
Xét trường hợp 1: \(x-2=4\)
\(\Rightarrow x=4+2\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-2=-4\)
\(\Rightarrow x=-4+2\)
\(\Rightarrow x=-\left(4-2\right)\)
\(\Rightarrow x=-2\)
Vậy \(x=6\) hoặc \(x=-2\)
b)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = 72
<=> 2x + 1 = 7
<=> x = 3
b. ( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = 34
<=> 2x - 1 = 3
<=> x = 2
c. ( x + 1 )3 = 2x3
<=> x + 1 = 2x
<=> x = 1
d. ( 2x + 1 )3 = 3x3
<=> 2x + 1 = 3x
<=> x = 1
( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = ( ±7 )2
<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = ( ±3 )4
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
( x + 1 )3 = ( 2x )3
<=> x + 1 = 2x
<=> x - 2x = -1
<=> -x = -1
<=> x = 1
( 2x + 1 )3 = ( 3x )3
<=> 2x + 1 = 3x
<=> 2x - 3x = -1
<=> -x = -1
<=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - 49 = 5 x 32
2x - 49 = 5 x 9
2x - 49 = 45
2x = 45+49=94
x = 94:2
x= 47
Vậy x=47
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