1) Ta có: \(4x^2-1=\left(2x+1\right).\left(3x-5\right)\)
\(\Leftrightarrow\left(2x+1\right).\left(2x-1\right)-\left(2x+1\right).\left(3x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right).\left[\left(2x-1\right)-\left(3x-5\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right).\left(2x-1-3x+5\right)=0\)
\(\Leftrightarrow\left(2x+1\right).\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\-x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=4\)
2) Ta có: \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-\left[2.\left(x-1\right)\right]^2=0\)
\(\Leftrightarrow\left[\left(x+1\right)+2.\left(x-1\right)\right].\left[\left(x+1\right)-2.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1+2x-2\right).\left(x+1-2x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right).\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)
Vậy \(x=\frac{1}{3}\) hoặc \(x=3\)
3) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x.\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-x+6x-3\right)=0\)
\(\Leftrightarrow x.\left[x.\left(2x-1\right)+3.\left(2x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x+3\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(TM\right)\\x=-\frac{1}{2}\left(TM\right)\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-3\) hoặc \(x=-\frac{1}{2}\)
4) Ta có: \(2x=3x-2\)
\(\Leftrightarrow2x-3x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\left(TM\right)\)
Vậy \(x=2\)
5) Ta có: \(x+15=3x-1\)
\(\Leftrightarrow x-3x=-1-15\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(TM\right)\)
Vậy \(x=8\)
6) Ta có: \(2-x=0,5x-4\)
\(\Leftrightarrow-x-0,5x=-4-2\)
\(\Leftrightarrow-1,5x=-6\)
\(\Leftrightarrow x=4\left(TM\right)\)
Vậy \(x=4\)
1) 4x2-1=(2x+1)(3x-5)
<=> (2x-1)(2x+1)-(2x+1)(3x-5)=0
<=> (2x+1)(2x-1-3x+5)=0
<=> (2x+1)(4-x)=0
<=>\([^{2x+1=0}_{4-x=0}< =>[^{2x=-1}_{x=4}< =>[^{x=\frac{-1}{2}}_{x=4}\)
2) (x+1)2= 4(x2-2x+1)
<=> x2+2x+1-4(x2-2x+1)=0
<=> x2+2x+1-4x2+8x-4=0
<=> -3x2+10x-3=0
<=> -3x2+x+9x-3=0
<=> -x(3x-1)+3(3x-1)=0
<=> (3x-1)(3-x)=0
<=> \([^{3x-1=0}_{3-x=0}< =>[^{3x=1}_{x=3}< =>[^{x=\frac{1}{3}}_{x=3}\)
3) 2x3+5x2-3x=0
<=> 2x(x2+\(\frac{5}{2}x-\frac{3}{2})=0\)
<=> 2x\(\left[x^2+2.\frac{5}{4}x+\frac{25}{16}-\left(\frac{25}{16}+\frac{3}{2}\right)\right]=0\)
<=> 2x\(\left[\left(x+\frac{5}{4}\right)^2-\frac{49}{16}\right]=0\)
<=> 2x\(\left(x+\frac{5}{4}-\frac{7}{4}\right)\left(x+\frac{5}{4}+\frac{7}{4}\right)=0\)
<=> x\(\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)
<=>\(\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)
4) 2x=3x-2
<=> 2x-3x=-2
<=> -x=-2
<=> x=2
5) x+15=3x-1
<=> x-3x=1-15
<=> -2x=-14
<=> x=-14:-2
<=> x=7
6) 2-x=0,5x-4
<=> -x-0,5x=-4-2
<=> -1,5x=-6
<=> x= -6: -1,5
<=> x=4
học tốt nghen
1, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
\(\Leftrightarrow4x^2-1=6x^2-10x+3x-5\)
\(\Leftrightarrow4x^2-1=6x^2-7x-5\)
\(\Leftrightarrow6x^2-7x-5-4x^2+1=0\)
\(\Leftrightarrow2x^2-7x-4=0\)
\(\Leftrightarrow\left(2x^2-8x\right)-\left(x-4\right)=0\)
\(\Leftrightarrow2x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{4;\frac{1}{2}\right\}\)
2, \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\\ \Leftrightarrow4x^2-8x+4-x^2-2x-1=0\\ \Leftrightarrow3x^2-10x+3=0\\ \Leftrightarrow\left(3x^2-x\right)-\left(9x-3\right)=0\\ \Leftrightarrow x\left(3x-1\right)-3\left(3x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{3;\frac{1}{3}\right\}\)
3,
\(2x^3+5x^2-3x=0\\ \Leftrightarrow x\left(2x^2+5x-3\right)=0\\ \Leftrightarrow x\left[\left(2x^2+x\right)-\left(6x+3\right)\right]=0\\ \Leftrightarrow x\left[x\left(2x+1\right)-3\left(2x+1\right)\right]=0\\ \Leftrightarrow x\left(2x+1\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x=0\\2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;-\frac{1}{2};3\right\}\)