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\(\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)
<=> \(\frac{60x-8-6\left(2x^2-x\right)}{12}\ge\frac{4x\left(1-3x\right)-15x}{12}\)
<=> \(60x-8-12x^2+6x\ge4x-12x^2-15x\)
<=> \(47x\ge8\)
<=> \(x\ge\frac{8}{47}\)
=>x^4+4x^2+9-4x^3-6x^2+12x<x^4-4x^3-2x^2+15x-3
=>-2x^2+12x+9<-2x^2+15x-3
=>-3x<-12
=>x>4
5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
Chúc bn học tốt❤
\(3\left(x-2\right)+4=5x-2\left(x-1\right)\\ \Leftrightarrow3x-6+4=5x-2x+2\\ \Leftrightarrow0x=4\left(vôlý\right)\)
Vậy pt vô nghiệm
\(2\left(x-2\right)-3\left(1-2x\right)=5\\ \Leftrightarrow2x-4-3+6x=5\\ \Leftrightarrow8x=12\\ \Leftrightarrow x=\dfrac{3}{2}\)
\(bpt\Leftrightarrow\left[\left(x+1\right)^2+3\right]\left(x-1\right)< 0\)
\(\left(x+1\right)^2+3>0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)
\(\Leftrightarrow x^4\left(x-1\right)-4x^3\left(x-1\right)+4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^4-4x^3+4x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x^3\left(x-1\right)-3x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^3-3x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)\left(x^2-4x+1\right)=0\)
- Khi x - 1 = 0 thì x = 1
- Khi x + 1 = 0 thì x = -1
- Khi \(x^2-4x+1=0\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}+2\\x=-\sqrt{3}+2\end{cases}}\)
Pt có tậo nghiệm là: \(S=\left\{1;-1;\sqrt{3}+2;-\sqrt{3}+2\right\}\)
