Bài 1 giải các phương trình
a) 2(x + 1) = 3 + 2x
b) 4x (1 - x) - 8 = 1 - (4x2 + 3)
c) x3 + 1 = x(x + 1)
d) 3x - 2/6 - 5 = 3 - 2(x + 7)/4
e) x2 -13x + 12 = 0
f) (2x - 1)2 - (x + 3)2 = 0
g) 2x/x - 1 + 4/x2 + 2x - 3 = 2x - 5/x + 3
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\(a)2\left(x+1\right)=3+2x\\ \Leftrightarrow2x+2=3+2x\\ \Leftrightarrow2x-2x=3-1\\ \Leftrightarrow0x=2\left(VN\right)\)
Vậy phương trình vô nghiệm
\(b)4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-8=-2\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(S=\left\{\dfrac{3}{2}\right\}\)
\(c)x^3+1=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1-x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{-1;1\right\}\)
\(d)\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow 12\left(\dfrac{3x-2}{6}-5\right)=12.\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow 6x-4-60=9-6\left(x+7\right)\)
\(\Leftrightarrow 6x-64=9-6x-42\)
\(\Leftrightarrow 6x-64=-6x-33\)
\(\Leftrightarrow 6x+6x=-33+64\\\Leftrightarrow 12x=31\\\Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(S=\left\{\dfrac{31}{12}\right\}\)
\(e)x^2-13x+12=0\\ \Leftrightarrow x^2-x-12x+12=0\\ \Leftrightarrow x\left(x-1\right)-12\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-12\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=12\end{matrix}\right.\)
Vậy \(S=\left\{1;12\right\}\)
\(f)\left(2x-1\right)^2-\left(x+3\right)^2=0\\ \Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Áp dụng hằng đẳng thức \(A^2-B^2=\left(A+B\right)\left(A-B\right)\)
