giải phương trình:
\(\frac{2}{x^2+3x+2}\)\(+\frac{1}{x^2+5x+6}\) \(=\frac{1}{x^2-3x+2}\)
Giải phương trình:
\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)
\(\Leftrightarrow\frac{x-5-x+1}{\left(x-1\right)\left(x-5\right)}=\frac{1}{8}\)
\(\Leftrightarrow-4.8=x^2-6x+5\)
\(\Leftrightarrow x^2-6x+37=0\)
Giải phương trình \(\left(\frac{7}{x^2+x-12}-\frac{1}{x^2-3x+2}-\frac{1}{x^2-5x+6}-\frac{3}{x^2+5x+4}\right)=\frac{3x}{x^2-1}\)
giải phương trình:
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
Để PT đc xác định : \(x^2+3x+2\ne0;x^2+5x+6\ne0;.....;x^2+15x+56\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\ne0;\left(x+2\right)\left(x+3\right)\ne0;....;\left(x+7\right)\left(x+8\right)\ne0\)
\(\Rightarrow x+1;x+2;x+3;....;x+8\ne0\)
\(\Rightarrow x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)
TXĐ : \(x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+....+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{7}{x^2+9x+8}=\frac{1}{14}\)
\(\Leftrightarrow x^2+9x+8=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)(TMĐKXĐ)
Vậy \(x=6\) hoặc \(x=-15\)
giải phương trình
\(\frac{3x-1}{2}-\frac{2-6x}{5}=\frac{1}{2}+\left(3x-1\right)\)
\(\left(x^2+2x+1\right)-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5x-5}{6}\)
\(\frac{3x-1}{2}-\frac{2-6x}{5}=\frac{1}{2}+\left(3x-1\right)\)
\(\Leftrightarrow\frac{3x-1}{2}+\frac{2\left(3x-1\right)}{5}-\left(3x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow\left(3x-1\right)\left(\frac{1}{2}+\frac{2}{5}-1\right)=\frac{1}{2}\)
\(\Leftrightarrow\frac{-1}{10}\left(3x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow3x-1=-5\)
\(\Leftrightarrow3x=-4\Leftrightarrow x=\frac{-4}{3}\)
Vậy nghiệm duy nhất của phương trình là\(x=\frac{-4}{3}\)
\(\left(x^2+2x+1\right)-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5x-5}{6}\)
\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5\left(x+1\right)}{6}\)
\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+6-5\right)}{6}\)
\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+1\right)}{6}\)
\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}-\frac{\left(x+1\right)\left(6x+1\right)}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-\frac{1}{3}-\frac{6x+1}{6}\right)=0\)
\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy nghiệm duy nhất của phương trình là\(x=-1\)
giải phương trình sau:
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\\\)
b) \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
c)\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)
d) \(5+\frac{76}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
b) Bạn kiểm tra lại đề bài
c) \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8}{16x^2-1}\left(x\ne\pm\frac{1}{4}\right)\)
\(\Leftrightarrow\frac{3}{1-4x}-\frac{2}{4x+1}+\frac{8}{16x^2-1}=0\)
\(\Leftrightarrow\frac{-3}{4x+1}-\frac{2}{4x+1}+\frac{8}{\left(4x+1\right)\left(4x-1\right)}=0\)
\(\Leftrightarrow\frac{-3\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}-\frac{2\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)
\(\Leftrightarrow\frac{-12x+3}{\left(4x-1\right)\left(4x+1\right)}-\frac{8x-2}{\left(4x-1\right)\left(4x+1\right)}+\frac{8}{\left(4x-1\right)\left(4x+1\right)}=0\)
\(\Leftrightarrow\frac{-12x+3-8x+2+8}{\left(4x-1\right)\left(4x+1\right)}=0\)
=> -20x+13=0
<=> -20x=-13
<=> \(x=\frac{13}{20}\left(tmđk\right)\)
Phương trình chứa ẩn ở mẫu
Giai các phương trình sau
1. \(\frac{7x-3}{x-1}=\frac{2}{3}\)
2. \(\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
3. \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
4. \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
5. \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
6. \(1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(1.\frac{7x-3}{x-1}=\frac{2}{3}\) ( \(x\ne1\))
\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)
\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\)
\(\Leftrightarrow x=\frac{7}{19}\)
\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)
\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)
\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)
\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)
\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)
\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-5\)
\(3.\frac{1-x}{x+1}+3=\frac{2x+3}{3x-1}\)
\(\Leftrightarrow\frac{\left(1-x\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}+\frac{3\left(x+1\right)\left(3x-1\right)}{\left(x+1\right)\left(3x-1\right)}=\frac{\left(2x+3\right)\left(x+1\right)}{\left(3x-1\right)\left(0+1\right)}\)
\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)
\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)
\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)
\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)
\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)
\(\Leftrightarrow4x^2+5x-7=0\)
\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)
\(\left(2x+\frac{5}{4}\right)^2>0\)
\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)
=> PT vô nghiệm
\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)
\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\)
\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)
\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow-6x=-16\)
\(\Leftrightarrow x=\frac{16}{6}\)
\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)
\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)
\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)
\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)
\(\Leftrightarrow x^4+x^3-4x-8=0\)
\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)
Đến đấy mk tắc r xl bạn nhé
Giải các phương trình:
\(a,\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(b,\frac{1}{x^2-5x+6}+\frac{2}{x^2-8x+15}+\frac{3}{x^2-13x+40}=\frac{6}{5}\)
Giải các phương trình sau:
a) \(\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\)
b) \(\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\)
c) \(\sin 3x - \cos 5x = 0\)
d) \({\cos ^2}x = \frac{1}{4}\)
e) \(\sin x - \sqrt 3 \cos x = 0\)
f) \(\sin x + \cos x = 0\)
a)
\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
b) \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
c)
\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)
d)
\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)
e)
\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)
f)
\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)
Bài 1. Giải các phương trình sau :
a) 7x - 35 = 0 b) 4x - x - 18 = 0
c) x - 6 = 8 - x d) 48 - 5x = 39 - 2x
Bài 2. Giải các phương trình sau :
a) 5x - 8 = 4x - 5 b) 4 - (x - 5) = 5(x - 3x)
c) 32 - 4(0,5y - 5) = 3y + 2 d) 2,5(y - 1) = 2,5y
Bài 3. Giải các phương trình sau :
a) \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
b) \(\frac{4x-7}{12}- x=\frac{3x}{8}\)
Bài 4. Giải các phương trình sau :
a) \(\frac{5x-8}{3}=\frac{1-3x}{2}\)
b) \(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
Bài 5. Giải các phương trình sau :
a) 6(x - 7) = 5(x + 2) + x b) 5x - 8 = 2(x - 4) + 3
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
có bị viết nhầm thì thông cảm nha!
la`thu'hai nga`y 19 nhe
\(\text{Giải phương trình sau:}\)
\(\frac{1}{x^2+1}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}=\frac{4}{5}\)
THKS~~~