a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
c)
\(\left(x^2-2x\right)\left(x^3-3x^2-18x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^3-6x^2+3x^2-18x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left[\left(x^3-6x^2\right)+\left(3x^2-18x\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left[x^2\left(x-6\right)+3x\left(x-6\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-6\right)\left(x^2+3x\right)=0\)
\(\Leftrightarrow x^2\left(x-6\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\\x=2\\x=-3\end{matrix}\right.\)
Mk làm luôn nhé , không chép lại đề đâu !
b) \(\dfrac{6}{\left(x-1\right)\left(x+1\right)}+5=\dfrac{8x-1}{4\left(x+1\right)}+\dfrac{12x-1}{4\left(x-1\right)}\) ( x # 1 , x # -1 )
<=> \(\dfrac{4\left(6+5x^2-5\right)}{4\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)\left(8x-1\right)}{4\left(x+1\right)\left(x-1\right)}+\dfrac{\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}\)
<=> 20x2 + 4 = 8x2 - 9x + 1 + 12x2 + 11x - 1
<=> 2x - 4 = 0
<=> x - 2 = 0
<=> x = 2 ( thỏa mãn ĐKXĐ )
Vậy,.....
