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Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\)
\(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\)
\(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \)
Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\)
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
6). ĐK x\(\ne\)...
PT \(\Leftrightarrow\dfrac{4}{2x^3+3x^2-8x-12}-\dfrac{1\left(2x+3\right)}{2x^3+3x^2-8x-12}-\dfrac{4\left(x-2\right)}{2x^3+3x^2-8x-12}+\dfrac{1\left(x^2-4\right)}{2x^3+3x^2-8x-12}=0\)(cái này bạn lấy 2x3+3x2-8x-12 chia cho các mẫu khác bằng phương pháp chia hoocner)
\(\Leftrightarrow4-\left(2x+3\right)-4\left(x-2\right)+\left(x^2-4\right)=0\\ \Leftrightarrow x^2-6x+5=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
a) \(\left(2x-1\right)^2-\left(3x+5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1-3x-5\right)=0\\ \text{}\Leftrightarrow\left(2x-1\right)\left(-x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-6\right\}\)
b) \(\dfrac{x+5}{4}-\dfrac{2x-3}{3}=\dfrac{2x-1}{12}\)
\(\Leftrightarrow3\left(x+5\right)-4\left(2x-3\right)=2x-1\\ \Leftrightarrow3x+15-8x+12=2x-1\\ \Leftrightarrow-5x+27=2x-1\\ \Leftrightarrow-5x-2x=-1-27\\ \Leftrightarrow-7x=-28\\ \Leftrightarrow x=4\)
Vậy \(S=\left\{4\right\}\)
\(c)\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{-4}{1-x^2}\)(ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{4}{x+1}=0\)
\(\Leftrightarrow4=0\)(vô lý)
Vậy .....
\(d)\dfrac{1}{x+1}+\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}\)(ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2}{x^2-x+1}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x-1-2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+2x-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x-2}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy ....
a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)
\(\Leftrightarrow15x-9x+6=45-10x+25\)
\(\Leftrightarrow15x-9x+10x=45+25-6\)
\(\Leftrightarrow16x=64\)
\(\Leftrightarrow x=4\)
b) \(x^2-9+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)
\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)
a) 15x - 3(3x - 2) = 45 - 5(2x - 5)
\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25
\(\Leftrightarrow\) 6x + 10x = 70 - 6
\(\Leftrightarrow\) 16x = 64
\(\Leftrightarrow\) x = 4
Vậy.......................
b) x2 - 9 + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)
Vậy........................
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4
\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0
\(\Leftrightarrow\) x2 - 6x = 0
\(\Leftrightarrow\) x(x - 6) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)
Vậy...............
\(a,3x-12=0\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4\)
\(b,\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-6\left(x-2\right)-x^2}{x^2-4}=0\)
\(\Leftrightarrow x^2+4x+4-6x+12-x^2=0\)
\(\Leftrightarrow-2x+16=0\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(tmdk\right)\)
\(a,3x-12=0\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4.\)
Vậy \(S=\left\{4\right\}\)
\(b,\left(x-2\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-3}{2}.\end{matrix}\right.\)
Vậy \(S=\left\{2;\dfrac{-3}{2}\right\}\)
\(c,\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow x^2+4x+4-6x+12-x^2=0\)
\(\Leftrightarrow-2x+16=0\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(tm\right).\)
Vậy \(S=\left\{8\right\}\)