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a,\(\Leftrightarrow9x^2+4x-3-9x^2-12x-4>0\)
\(\Leftrightarrow-8x-7>0\)
\(\Leftrightarrow-8x>7\)\(\Leftrightarrow x< -\dfrac{7}{8}\)
\(b,\Leftrightarrow\dfrac{4x^2-2\left(2x^2+3x\right)}{4}< \dfrac{x-1}{4}\)
\(\Leftrightarrow4x^2-4x^2-6x< x-1\)
\(\Leftrightarrow-6x-x< x-1\)
\(\Leftrightarrow-7x< -1\Leftrightarrow x>\dfrac{1}{7}\)
Vậy....
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)
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 \(x=\dfrac{17}{37}\)
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)
Suy ra: \(3x^2+7x-10=0\)
\(\Leftrightarrow3x^2-3x+10x-10=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)
a/ \(\dfrac{3x^2+7x-10}{x}=0\)
\(< =>3x^2+7x-10=0\)
\(< =>3x^2+10x-3x-10=0\)
\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)
\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)
\(< =>\left(3x+10\right)\left(x-1\right)=0\)
\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)
Vậy tập nghiệm của .....
a: \(\dfrac{2x-6}{x+2}>0\)
=>x-3>0 hoặc x+2<0
=>x>3 hoặc x<-2
b:
Theo BXD, ta có: f(x)>0
=>-3<x<1 hoặc x>2