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\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
a)
\(\frac{1-4x}{12}< \frac{5-3x}{9}\Rightarrow9-36x< 60-36x\)
\(\Rightarrow9< 60\)
=> \(\forall x\in R\)
b)
\(\frac{x-1}{x-2}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
x thỏa mãn là : x < 1 hoặc x >2
c) \(\frac{x+9}{x-1}>5\Rightarrow x+1-5\left(x-1\right)>0\Rightarrow x+1-5x+5>0\)
\(\Rightarrow6-4x>0\Rightarrow4x< 6\Rightarrow x< \frac{2}{3}\)
d) Tương tự câu c
e) \(\left(x-1\right)\left(x+2\right)< \left(x+4\right)^2-4\)
\(\Rightarrow x^2+x-2< x^2+8x+16-4\Rightarrow-7x-14< 0\)
\(\Rightarrow7x+14>0\Rightarrow x>-2\)
c) \(\left|2x-3\right|=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
\(TH:2x-3=4\)
\(\Leftrightarrow2x=4+3\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\frac{7}{2}\)
\(TH:2x-3=-4\)
\(\Leftrightarrow2x=-4+3\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(x\in\left\{\frac{7}{2};\frac{-1}{2}\right\}\)
e) \(\frac{x-1}{x-3}>1\)
\(ĐKXĐ:x\ne3\)
\(\Leftrightarrow\frac{x-3+2}{x-3}>1\)
\(\Leftrightarrow\frac{x-3}{x-3}+\frac{2}{x-3}>1\)
\(\Leftrightarrow1+\frac{2}{x-3}>1\)
\(\Leftrightarrow\frac{2}{x-3}>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)