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\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
a) \(6x^2-15x\)
b) \(x^2+5x+4\)
c) \(49-x^2\)
d) \(x^2+4x+4\)
e) \(9-12x+4x^2\)
f) \(x^3-8\)
\(a,=6x^2-15x\\ b,=x^2+5x+4\\ c,=49-x^2\\ d,=x^2+4x+4\\ e,=9-12x+4x^2\\ f,=x^3-8\)
\(\left(2x-5\right)\left(x-3\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(x-3+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(3x-8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\3x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{8}{3}\end{cases}}\)
\(\frac{3x-5}{4}+\frac{2x-3}{6}=\frac{x}{3}-1\)
\(\Leftrightarrow\frac{18x-30+8x-12}{24}=\frac{x-3}{3}\)
\(\Leftrightarrow\frac{26x-42}{24}=\frac{x-3}{3}\)
\(\Leftrightarrow78x-126=24x-72\)
Chuyển vế các kiểu
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
1) \(3x\left(x-4\right)-x+4=0\)
\(\Rightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
2) \(2x\left(2x+3\right)-2x-3=0\)
\(\Rightarrow2x\left(2x+3\right)-\left(2x+3\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(3x\left(x-4\right)-x+4=0\\ \Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\\ 2x\left(2x+3\right)-2x-3=0\\ \Leftrightarrow\left(2x+3\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)