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5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
0 8/3
Chúc bn học tốt❤
\(\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)
<=> \(\frac{60x-8-6\left(2x^2-x\right)}{12}\ge\frac{4x\left(1-3x\right)-15x}{12}\)
<=> \(60x-8-12x^2+6x\ge4x-12x^2-15x\)
<=> \(47x\ge8\)
<=> \(x\ge\frac{8}{47}\)
Ix-1I+Ix-2I>x+3 (1)
Ta xét các TH về giá trị của x:
TH1: \(x< -1\)
(1) \(\leftrightarrow1-x+2-x>x+3\)
\(\leftrightarrow3-x>x+3\)
\(\leftrightarrow x< 0\) (2)
TH2:\(-1\le x< 2\)
(1)\(\leftrightarrow x-1+2-x>x+3\)
\(\leftrightarrow1>x+3\)
\(\leftrightarrow x< -2\)(loại) (3)
TH3:\(x\ge2\)
(1)\(\leftrightarrow x-1+x-2>x+3\)
\(\leftrightarrow2x-3>x+3\)
\(\leftrightarrow x>6\) (4)
Từ (2),(3) và (4) \(\rightarrow\orbr{\begin{cases}x< 0\\x>6\end{cases}}\)
giai di giai di giai di............................................................
giai di ma , lam on
- Ta có: \(\left(x^2-1\right).\left(x+2\right).\left(x-3\right)=\left(x-1\right).\left(x^2-4\right).\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x+2\right).\left(x-3\right)=\left(x-1\right).\left(x-2\right).\left(x+2\right).\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x+2\right).\left(x-3\right)-\left(x-1\right).\left(x-2\right).\left(x+2\right).\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left[\left(x+1\right).\left(x-3\right)-\left(x-2\right).\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left[\left(x^2-2x-3\right)-\left(x^2+3x-10\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x+2\right).\left(-5x+7\right)=0\)
+ \(x-1=0\)\(\Leftrightarrow\)\(x=1\left(TM\right)\)
+ \(x+2=0\)\(\Leftrightarrow\)\(x=-2\left(TM\right)\)
+ \(-5x+7=0\)\(\Leftrightarrow\)\(-5x=-7\)\(\Leftrightarrow\)\(x=\frac{7}{5}\left(TM\right)\)
Vậy \(S=\left\{-2,1,\frac{7}{5}\right\}\)
1.: Áp dụng BĐT Cauchy-Schwarz cho 3 số dương
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)