1Giải các phương trình sau bằng cách đưa về phương trình tích:
a) 3x(2x – 3) = 5(3 – 2x)
b) (x2 + 1)(2x + 5) = (x – 1)(x2 + 1)
c) 3x3 = x2 + 3x - 1
d) x2 – 9x + 20 = 0
2Giải các phương trình sau bằng cách đưa về phương trình tích:
a) 3x(2x – 3) = 5(3 – 2x)
b) (x2 + 1)(2x + 5) = (x – 1)(x2 + 1)
c) 3x3 = x2 + 3x - 1
d) x2 – 9x + 20 = 0
a/ \(3x(2x-3)=5(3-2x) \Leftrightarrow 3x(2x-3)+5(2x-3)=0 \\\ \Leftrightarrow (2x-3)(3x+5)=0 \)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{5}{3}\end{matrix}\right.\)
KL: .............
b/ \(\left(x^2+1\right)\left(2x+5\right)=\left(x-1\right)\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\left(2x+5\right)-\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+5-x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-6\end{matrix}\right.\)
KL: .............
c/ \(3x^3=x^2+3x-1\Leftrightarrow3x^3-x^2-3x+1=0\Leftrightarrow x^2\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=1\\x=-1\end{matrix}\right.\)
KL: ..........
d/ \(x^2-9x+20=0\Leftrightarrow x^2-5x-4x+20=0\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
KL: .............
\(a,3x\left(2x-3\right)=5\left(3-2x\right)\Leftrightarrow6x^2-9x=15-10x\Leftrightarrow6x^2-9x-15+10x=0\Leftrightarrow6x^2+x-15=0\Leftrightarrow\left(3x+5\right)\left(2x-3\right)=0\)
\(\left(3x+5\right)\left(2x-3\right)=0\)
\(\left[{}\begin{matrix}x=-\frac{5}{3}\\x=\frac{3}{2}\end{matrix}\right.\)
\(\left(x^2+1\right)\left(2x+5\right)=\left(x+1\right)\left(x^2+1\right)\)
\(x^3+6x^2+x+6=0\)
\(\left(x+6\right)\left(x^2+1\right)=0\)
\(\left[{}\begin{matrix}x=-6\\x=\pm\sqrt{-1}\end{matrix}\right.\)
\(3x^3=x^2+3x-1\)
\(3x^3-x^2-3x+1=0\)
\(\left(3x-1\right)\left(x^2-1\right)=0\)
\(\left[{}\begin{matrix}3x=1\\x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{1}{3}\\x=\pm1\end{matrix}\right.\)
phần b mk nhầm nha , x=-6 thôi tại vì x bình phương là dương mà -1 là âm nên vô lí nha !
\(x^2-9x+20=0\)
\(\left(x-5\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
