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a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
\(\frac{x-5}{3}< \frac{x-8}{4}\Rightarrow4.\left(x-5\right)< 3.\left(x-8\right)\Rightarrow4x-20< 3x-24\Rightarrow x< -4\)
a) \(\frac{x-5}{3}< \frac{x-8}{4}\)
<=> \(\frac{4\left(x-5\right)}{12}< \frac{3\left(x-8\right)}{12}\)
<=> \(4\left(x-5\right)< 3\left(x-8\right)\)
<=> \(4x-20< 3x-24\)
<=> \(4x-3x< 20-24\)
<=> \(x< -4\)
Vậy bất phương trình có tập nghiệm là { x l x < -4 }
b) \(\frac{x+3}{4}+1< x+\frac{x+2}{3} \)
<=> \(\frac{3\left(x+3\right)}{12}+\frac{12}{12}< \frac{12x}{12}+\frac{4\left(x+2\right)}{12}\)
<=> \(3\left(x+3\right)+12< 12x+4\left(x+2\right)\)
<=> \(3x+9+12< 12x+4x+8\)
<=> \(3x-12x-4x< 8-9-12\)
<=> \(-13x< -13\)
<=> \(x>1\)
Vậy bất phương trình có tập nghiệm là { x l x > 1 }
a, Đặt \(x^2-4x+8=a\left(a>0\right)\)
\(\Rightarrow a-2=\frac{21}{a+2}\)
\(\Leftrightarrow a^2-4=21\Rightarrow a^2=25\Rightarrow a=5\)
Thay vào là ra
b) ĐK: \(y\ne1\)
bpt <=> \(\frac{4\left(1-y\right)}{1-y^3}+\frac{1+y+y^2}{1-y^3}+\frac{2y^2-5}{1-y^3}\le0\)
<=> \(\frac{3y^2-3y}{1-y^3}\le0\)
\(\Leftrightarrow\frac{y\left(y-1\right)}{\left(y-1\right)\left(y^2+y+1\right)}\ge0\)
\(\Leftrightarrow\frac{y}{y^2+y+1}\ge0\)
vì \(y^2+y+1=\left(y+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
nên bpt <=> \(y\ge0\)
1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
ta có:
\(\frac{x+2}{2013}+\frac{x+5}{2010}>\frac{x+8}{2007}+\frac{x+11}{2004}\)
\(\Leftrightarrow\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+5}{2010}+1\right)>\left(\frac{x+8}{2007}+1\right)+\left(\frac{x+11}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}>\frac{x+2015}{2007}+\frac{x+2015}{2004}\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}-\frac{x+2015}{2007}-\frac{x+2015}{2004}>0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}\right)>0\)
\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)
\(a,4x-6< 7x-12\)
\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)
\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)
\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)
\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)
\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)
\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)
\(\Leftrightarrow-16x\ge11\)
\(\Leftrightarrow x\le-\frac{11}{16}\)
\(d,-12-8x>3+2x-\left(5-7x\right)\)
\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)
\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)
\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)
\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)
a) \(4x+5\ge7\)
\(\Leftrightarrow4x\ge3\)
\(\Leftrightarrow x\ge\frac{3}{4}\)
Vậy BPT có nghiệm \(x\ge\frac{3}{4}\)
b) \(\frac{x-1}{4}-1>\frac{x+1}{3}+8\)
\(\Leftrightarrow\frac{x-1}{4}-\frac{x+1}{3}>8+1\)
\(\Leftrightarrow\frac{3\left(x-1\right)}{3.4}-\frac{4\left(x+1\right)}{4.3}>9\)
\(\Leftrightarrow\frac{3x-3}{12}-\frac{4x+4}{12}>9\)
\(\Leftrightarrow\frac{3x-3-4x-4}{12}>9\)
\(\Leftrightarrow-x-7>108\)
\(\Leftrightarrow-7-108>x\)
<=> -115>x
Vậy BPT có nghiệm là x<-115