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\(5x-\frac{3-2x}{2}>\frac{7x-5}{2}+x\)
\(\Leftrightarrow\) \(\frac{10x}{2}-\frac{3-2x}{2}>\frac{7x-5}{2}+\frac{2x}{2}\)
\(\Rightarrow\) \(10x-3+2x>7x-5+2x\)
\(\Leftrightarrow\) \(10x+2x-7x-2x>-5+3\)
\(\Leftrightarrow\) \(3x>-2\)
\(\Leftrightarrow\) \(x>-\frac{2}{3}\)
Vậy ................
\(\Leftrightarrow\dfrac{7x-8}{32}-\dfrac{2\left(5-x\right)}{32}>\dfrac{16\left(x+9\right)}{32}+\dfrac{4}{32}\)
\(\Leftrightarrow7x-8-2\left(5-x\right)>16\left(x+9\right)+4\)
\(\Leftrightarrow7x-8-10+2x>16x+148\)
\(\Leftrightarrow-7x>166\)
\(\Rightarrow x< -\dfrac{166}{7}\)
Ta có : \(\dfrac{3-7x}{1+x}\ge\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3-7x}{1+x}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{2\left(3-7x\right)-\left(x+1\right)}{2\left(x+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{5-15x}{2\left(x+1\right)}=\dfrac{5\left(3-x\right)}{2\left(x+1\right)}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x\ge0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x\le0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le3\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge3\\x< -1\end{matrix}\right.\end{matrix}\right.\)
Vậy suy ra tập nghiệm
b, (x+4)(5x+9)-x>4
\(\Leftrightarrow\)5x2+29x+36-x>4
\(\Leftrightarrow\)5x2+28x+36>4
\(\Leftrightarrow\)5x2+28x+32>0
\(\Leftrightarrow\)5(x2+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\))>0
\(\Leftrightarrow\)x2+\(\dfrac{28}{5}\)x+\(\dfrac{32}{5}\)>0
\(\Leftrightarrow\)x2+2.\(\dfrac{14}{5}\)x+\(\dfrac{206}{25}\)+\(\dfrac{32}{5}\)-\(\dfrac{206}{25}\)>0
\(\Leftrightarrow\)(x+\(\dfrac{14}{5}\))2-\(\dfrac{46}{25}\)>0
\(\Leftrightarrow\)(x+\(\dfrac{14-\sqrt{46}}{5}\))(x+\(\dfrac{14+\sqrt{46}}{5}\))>0
\(\Leftrightarrow\)2 trường hợp
pt <=> 8x+12-7x+7=40x-8+166
<=> x+19=40x+158
<=> 39x=-139
<=> x=-139/39
a) 3(x-2)(x+2) < 3x2 + x
3(x2 + 2x - 2x - 4 ) < 3x2 + x
<=> 3x2 + 6x - 6x - 12 < 3x2 + x
<=> 3x2 + 6x - 6x - 3x2 - x < 12
<=> x > -12
Vậy bpt có nghiệm là x > -12.
b) ( x+4 )(5x-1) > 5x2 + 16x + 2
<=> 5x2 - x + 20x - 4 - 5x2 - 16x - 2 > 0
<=> 5x2 - x + 20x - 5x2 - 16x > 2 + 4
<=> 3x > 6
<=> x > 2
Vậy btp có nghiệm là x > 2
Giải:
a) \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
\(\Leftrightarrow3\left(x^2-4\right)< 3x^2+x\)
\(\Leftrightarrow3x^2-12< 3x^2+x\)
\(\Leftrightarrow-12< x\)
\(\Leftrightarrow x>-12\)
Vậy ...
b) \(\left(x+4\right)\left(5x-1\right)>5x^2+16x+2\)
\(\Leftrightarrow5x^2+20x-x-4>5x^2+16x+2\)
\(\Leftrightarrow5x^2+19x-4>5x^2+16x+2\)
\(\Leftrightarrow3x-4>2\)
\(\Leftrightarrow3x>6\)
\(\Leftrightarrow x>2\)
Vậy ...
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)
\(10x-15-20x+28=19-2x-22\)
\(10x-20x+2x=19-22-28+15\)
\(-8x=-16\)
\(\Rightarrow x=2\)
\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)
\(4x+12-7x+17=40x-8+166\)
\(4x-7x-40x=-8+166-17-12\)
\(-43x=129\)
\(x=-3\)
\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)
\(17-14x+14=13-4x-4-5x+15\)
\(-14x+4x+5x=13-4+15-14-17\)
\(-5x=-7\)
\(x=\frac{7}{5}\)
\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)
\(5x+3,5+3x-4=7x-3x+1,5\)
\(5x+3x-7x+3x=1,5-3,5\)
\(x=-2\)
\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)
\(28x+21-4x+4=15x+11,25+7\)
\(28x-4x-15x=11,25+7-4-21\)
\(9x=\frac{-27}{4}\)
\(x=\frac{-3}{4}\)
\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)
\(3x+0,8x+0,2x=3,38-2,42\)
\(4x=\frac{24}{25}\)
\(x=\frac{6}{25}\)
chúc bạn học tốt !!
\(7x+4\ge5x-8\)
\(\Leftrightarrow7x-5x=-4-8\)
\(\Leftrightarrow2x\ge-12\)
\(\Leftrightarrow x\ge-6\)
S=\(\left\{x/x\ge-6\right\}\)