Tìm x, biết:
a) (3x - 2)2 - x(9x - 4) = 0
<=> 9x2 - 12x + 4 - 9x2 + 4x = 0
<=> -8x + 4 = 0
<=> -8x = -4
<=> x = \(\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b) x(7 - x) + (x + 2)2 = 15
<=> 7x - x2 + x2 + 4x + 4 = 15
<=> 11x + 4 = 15
<=> 11x = 11
<=> x = 1
Vậy x = 1
c) (2x - 3)2 - 2(x - 1)(2x - 3) + (x - 1)2 = 9
<=> 4x2 - 12x + 9 - 2(2x2 - 3x - 2x + 3) + x2 - 2x + 1 = 9
<=> 4x2 - 12x + 9 - 4x2 + 6x + 4x - 6 + x2 - 2x + 1 = 9
<=> x2 - 4x + 4 = 9
<=> x2 - 4x - 5 = 0
<=> x2 - 5x + x - 5 = 0
<=> x(x - 5) + (x - 5) = 0
<=> (x - 5)(x + 1) = 0
<=> \(\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy x ={5; -1}
a) \(\left(3x-2\right)^2-x\left(9x-4\right)=0\)
\(\Rightarrow\left(3x-2\right)^2=x\left(9x-4\right)\)
\(\Rightarrow\left[\left(3x\right)^2-2.3x.2+2^2\right]=9x^2-4x\)
\(\Rightarrow9x^2-12x+4=9x^2-4x\)
\(\Rightarrow9x^2-9x^2-12x+4x=-4\)
\(\Rightarrow-8x=-4\)
\(\Rightarrow x=\frac{-4}{-8}=\frac{1}{2}\)
b) \(x\left(7-x\right)+\left(x+2\right)^2=15\)
\(\Rightarrow7x-x^2+x^2+4x+4=15\)
\(\Rightarrow11x=15-4=11\)
\(\Rightarrow x=11:11=1\)
c)\(\left(2x-3\right)^2-2\left(x-1\right)\left(2x-3\right)+\left(x-1\right)^2=9\)
\(\Rightarrow4x^2-12x+9-\left(2x-2\right)\left(2x-3\right)+\left(x^2-2x+1\right)=9\)
\(\Rightarrow4x^2-12x+9-\left[2x\left(2x-3\right)-2\left(2x-3\right)\right]+x^2-2x+1=9\)
\(\Rightarrow4x^2-12x+9-\left(4x^2-6x-4x+6\right)+x^2-2x+1=9\)
\(\Rightarrow4x^2-12x+9-4x^2+6x+4x-6+x^2-2x+1=9\)
\(\Rightarrow-4x+x^2=9-9+6-1=5\)
\(\Rightarrow x\left(-4+x\right)=5\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=5\\-4+x=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\-4+x=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\-4+x=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5\\-4+x=-1\end{matrix}\right.\end{matrix}\right.\)
Ta có bảng sau:
|
X = 1 |
X = 5 |
X = -1 |
X = -5 |
|
-4 + x = 5 => x = 5 + 4 = 9 |
-4 + x = 1 => x = 1 + 4 = 5 |
-4 + x = -5 => x = -5 + 4 = -1 |
-4 + x = -1 => x = -1 + 4 = 3 |
|
Loại |
Chọn |
Chọn |
Loại |
