|x - \(\frac{1}{2}\)| . ( x + 1 ) = 0
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a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\) TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\) TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)
\(\frac{1}{7}x=\frac{2}{7}\) \(-\frac{1}{5}x=\frac{3}{5}\) \(\frac{1}{3}x=\frac{4}{3}\)
\(x=\frac{2}{7}\cdot7\) \(x=\frac{3}{5}\cdot-5\) \(x=\frac{4}{3}\cdot3\)
\(x=2\) \(x=-3\) \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)
\(x\cdot\frac{5+3-24}{30}=1\)
\(x\cdot\frac{-8}{15}=1\)
\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)
a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x2 - 3x +1>0
<=> 2x2 - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
\(|x-\frac{1}{2}|.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}|x-\frac{1}{2}|=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
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