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2.a)
\(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)
\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)
\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\)
\(\Leftrightarrow-3x>6\)
\(\Leftrightarrow3>\dfrac{6}{-3}\)
\(\Leftrightarrow x< -2\)
Vậy nghiệm của bpt \(S=\left\{-2\right\}\)
2.b)
\(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\)
\(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\)
\(\Leftrightarrow4x+4-12\ge3x-6\)
\(\Leftrightarrow4x-3x\ge-6-4+12\)
\(\Leftrightarrow x\ge2\)
vậy nghiệm của bpt x\(\ge\)2
c: \(\Leftrightarrow2x-8>=2x+1\)
=>-8>=1(vô lý)
d: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>10x-30>3x+5
=>7x>35
hay x>5
\(a,2x+7\ge0\Leftrightarrow2x\ge-7\Rightarrow x\ge\dfrac{-7}{2}\)
\(b,5-2x\le0\Leftrightarrow-2x\le-5\Leftrightarrow x\ge\dfrac{5}{2}\)
\(c,\dfrac{x+2}{x^2+1}\ge0\Leftrightarrow x+2\ge x^2+1\Leftrightarrow x+2-x^2-1\ge0\Leftrightarrow x-x^2+1\ge0\)\(\Leftrightarrow-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\ge0\Leftrightarrow-\left(x-\dfrac{1}{2}\right)^2\ge-\dfrac{5}{4}\Rightarrow\left(x-\dfrac{1}{2}\right)^2\ge\dfrac{5}{4}\)\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\ge\sqrt{\dfrac{5}{4}}\\x-\dfrac{1}{2}\ge-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\\x\ge-\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\end{matrix}\right.\)
\(d,\dfrac{x^2+3}{2-x}< 0\Leftrightarrow x^2+3< 2-x\Leftrightarrow x^2+3-2+x\ge0\Leftrightarrow\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2\ge\dfrac{-3}{4}\)( vô lí )
Vậy : BPT trên vô nghiệm
\(\dfrac{-4x-1}{3}-\dfrac{2-x}{15}\ge\dfrac{2x-3}{5}\)
\(\Leftrightarrow\) \(\dfrac{5\left(-4x-1\right)}{15}-\dfrac{2-x}{15}\ge\dfrac{3\left(2x-3\right)}{15}\)
\(\Leftrightarrow\) -20x - 5 - 2 + x \(\ge\) 6x - 9
\(\Leftrightarrow\) -19x - 7 \(\ge\) 6x - 9
\(\Leftrightarrow\) -19x - 6x \(\ge\) -9 + 7
\(\Leftrightarrow\) -25x \(\ge\) -2
\(\Leftrightarrow\) x \(\le\) \(\dfrac{2}{25}\)
\(\dfrac{-4x-1}{3}-\dfrac{2-x}{15}\) ≥\(\dfrac{2x-3}{5}\)
⇔ -5(4x+1)-2-x≥3(2x-3)
⇔ -21x-7 ≥ 6x-9
⇔-21x-6x ≥ 7-9
⇔ -27x ≥ -2
⇔ x ≤ 2/27
0 2/27
hình hơi xấu và mgang tính chất minh họa nên bạn thông cảm 
\(\dfrac{x}{3}+\dfrac{2x-4}{4}\ge\dfrac{x}{6}+x\)
\(\dfrac{2x}{6}+\dfrac{x}{2}-1\ge\dfrac{x}{6}+\dfrac{2x}{2}\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{x}{6}+\dfrac{x}{2}-\dfrac{2x}{2}\ge1\)
\(\Leftrightarrow\dfrac{x}{6}-\dfrac{x}{2}\ge1\)
\(\Leftrightarrow\dfrac{x-3x}{6}\ge1\)
\(\Leftrightarrow-2x\ge6\)
\(\Leftrightarrow x\ge-3\)
Vậy BPT có tập nghiệm là ;\(S=\left\{x\ge-3\right\}\)
a. \(5x-10=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b. \(\dfrac{2}{x+1}-\dfrac{3}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne-1;x\ne2\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{3\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2\left(x-2\right)-3\left(x+1\right)=2x-6\)
\(\Leftrightarrow2x-4-3x-3=2x-6\)
\(\Leftrightarrow2x-4-3x-3-2x+6=0\)
\(\Leftrightarrow-3x-1=0\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
c) \(3x-5\ge-7\) (3)
\(\Leftrightarrow3x\ge-7+5\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
Vậy tập nghiệm của BPT (3) là \(x\ge-\dfrac{2}{3}\)
d) \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(S=\left\{\dfrac{1}{3}\right\}\)
a. 5x - 10 = 0
⇔ 5x = 10
⇔ x = 2
Vậy S ={2}.
b.\(\dfrac{2}{x+1}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔\(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\) - \(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔ 2(x - 2) - 3(x+1) = 2x - 6
⇔ 2x - 4 - 3x -3 = 2x - 6
⇔ 2x - 2x - 3x = -6 + 4 + 3
⇔ -3x = 1
⇔ x = \(-\dfrac{1}{3}\)
\(\dfrac{3x-1}{5}\ge\dfrac{x}{2}+0,8\left(1\right)\\ 1-\dfrac{2x-5}{6}>\dfrac{3-x}{4}\left(2\right)\)
+) Giải \(\left(1\right):\dfrac{3x-1}{5}\ge\dfrac{x}{2}+0,8\)
\(\Leftrightarrow2\left(3x-1\right)\ge5x+8\\ \Leftrightarrow6x-2\ge5x+8\\ \Leftrightarrow6x-5x\ge8+2\\ \Leftrightarrow x\ge10\)
+) Giải \(\left(2\right):\) \(1-\dfrac{2x-5}{6}>\dfrac{3-x}{4}\)
\(\Leftrightarrow12-2\left(2x-5\right)>3\left(3-x\right)\\ \Leftrightarrow12-4x+10>9-3x\\ \Leftrightarrow-4x+3x>9-22\\ \Leftrightarrow-x>-11\\ \Leftrightarrow x< 11\)
Vậy \(10\le x< 11\) thỏa mãn cả 2 bất phương trình
a ) \(5-7x\ge2x+14\)
\(\Leftrightarrow-7x-2x\ge14-5\)
\(\Leftrightarrow-9x\ge9\)
\(\Leftrightarrow x\le-1\)
Vậy bất phương trình có nghiệm \(x\le1\)
b ) \(\dfrac{1,5-x}{5}< \dfrac{4x+5}{2}\)
\(\Leftrightarrow2\left(1,5-x\right)< 5\left(4x+5\right)\)
\(\Leftrightarrow3-2x< 20x+25\)
\(\Leftrightarrow-2x-20x< 25-3\)
\(\Leftrightarrow-22x< 22\)
\(\Leftrightarrow x>-1\)
Vậy bất phương trình có nghiệm \(x>-1\)
Tick nha
a) 5 - 7x \(\ge\) 2x + 14
\(\Leftrightarrow\) -7x -2x \(\ge\) -5 + 14
\(\Leftrightarrow\) -9x \(\ge\) 9
\(\Leftrightarrow\) x \(\le\) -1
b) \(\dfrac{1,5-x}{5}< \dfrac{4x+5}{2}\)
\(\Leftrightarrow\) \(2\left(1,5-x\right)\) \(< 5\left(4x+5\right)\)
\(\Leftrightarrow\) \(3-2x\)\(< 20x+25\)
\(\Leftrightarrow\) \(-2x-20x< -3+25\)
\(\Leftrightarrow\) \(-22x< 22\)
\(\Leftrightarrow\) \(x>-1\)