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\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
1) 2x(x + 1) - x2(x + 2) + x3 - x + 4 = 0
<=> 2x.x + 2x.1 + (-x2).x + (-x2).2 + x3 - x + 4 = 0
<=> 2x2 + 2x - x3 - 2x2 + x3 - x = 0 - 4
<=> x = -4
=> x = -4
2) xem lại đề rồi chúng mình nói chuyện cậu nha :))
3) tương tự (mình hơi lười, thông cảm :v)
3, [(3x - 5)(7 - 5x)] - [(5x + 2)(2 - 3x)] = 4
<=> ( 21x -15x^2 -35 +25x) - (10x -15x^2 + 4-6x)=4
<=> 21x -15x^2 -35 +25x- 10x + 15x^2 - 4+6x =4
<=> 42x - 39 =4
<=> 42x = 43
<=< x =43/42
2, (3x - 2)(4x - 5 ) - (2x - 1)(6x + 2) = 0
12x2- 15x - 8x + 10 - 12x2 - 4x + 6x + 2 = 0
- 21x = -12
x = 4/7
1, đã có người giải
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(b,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow-\left(3x-2\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-11-2+5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)
\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)
\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-x^2-4x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)
\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-3\right\}\)
\(h,2x\left(x-1\right)=x^2-1\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(S=\left\{1\right\}\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
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