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\(x-4\)
\(\left(\sqrt{2}\right)^2-4\)
\(=\left(\sqrt{2}-2\right)\left(\sqrt{2}+2\right)\)
Xét \(5-x=0\Leftrightarrow x=5\)
\(x-1=0\Leftrightarrow x=1\)
\(2+3x=0\Leftrightarrow x=-\dfrac{2}{3}\)
Bảng xét dấu:
x -vc -2/3 1 5 +vc 5-x x-1 2+3x VT 0 0 0 0 0 0 - + + + - + + + + + - - - + - +
Để VT\(\le\)0 <=>\(\left[{}\begin{matrix}-\dfrac{2}{3}\le x\le1\\x\ge5\end{matrix}\right.\)
Vậy...
\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)
\(\Leftrightarrow2+x+1\le\dfrac{12}{4}-\dfrac{x-1}{4}\)
\(\Leftrightarrow x+3\le\dfrac{13-x}{4}\)
\(\Leftrightarrow\dfrac{4x+12}{4}\le\dfrac{13-x}{4}\)
\(\Leftrightarrow4x+12\le13-x\)
\(\Leftrightarrow4x+x\le13-12\)
\(\Leftrightarrow5x\le1\)
\(\Leftrightarrow x\le\dfrac{1}{5}\)
Vậy: \(x\le\dfrac{1}{5}\)
\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)
\(\Leftrightarrow\dfrac{12x+36}{12}\le\dfrac{33-3x}{12}\)
\(\Leftrightarrow12x+36\le33-3x\)
\(\Leftrightarrow12x+3x\le-36+33\)
\(\Leftrightarrow15x\le-3\)
\(\Leftrightarrow x\le\dfrac{-1}{5}\)
\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)
\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)
\(\Leftrightarrow4x-2< 3x+18\)
\(\Leftrightarrow4x-3x< 2+18\)
\(\Leftrightarrow x< 20\)
\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)
\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)
\(\Leftrightarrow5x-11>4x+4\)
\(\Leftrightarrow5x-4x>11+4\)
\(\Leftrightarrow x>15\)
mk lm k chắc đúng, sai đâu ib mk nhé
DKXD: \(x\ge-\frac{1}{2};\)\(x\ne0\)
Dat: \(\sqrt{2x+1}=a\) \(\left(a\ge0;a\ne1\right)\)
Khi đó bpt đã cho trở thành:
\(\frac{a^2-1}{a-1}>a^2+1\)
<=> \(a+1>a^2+1\)
<=> \(a\left(1-a\right)>0\)
<=> \(1-a>0\)
<=> \(a< 1\)
Khi đó: \(\sqrt{2x+1}< 1\)
<=> \(2x+1< 1\)
<=> \(x< 0\)
Vay: \(-\frac{1}{2}\le x< 0\)