a/\(\left(x+3\right)\left(\frac{40}{x}+3\right)=88\)(đk: x\(\ne0\))
\(\Leftrightarrow40+\frac{120}{x}+3x+9=88\)
\(\Leftrightarrow3x+\frac{120}{x}=39\) \(\Leftrightarrow\frac{3x^2+120}{x}=39\)
\(\Leftrightarrow x^2+40-13x=0\Leftrightarrow\left(x-8\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)(tm)
vậy x=8 hoặc x=5 là nghiệm của phương trình
b/ \(\frac{x-1}{2x+1}< \frac{1}{2}\left(x\ne-\frac{1}{2}\right)\)
\(\Leftrightarrow\frac{x-1}{2x+1}-\frac{1}{2}< 0\) \(\Leftrightarrow\frac{2x-2}{4x+2}-\frac{2x+1}{4x+2}< 0\)
\(\Leftrightarrow\frac{-3}{4x+2}< 0\) \(\Leftrightarrow4x+2>0\)(vì -3<0)
<=> x>-1/2(tm)
vậy x>-1/2 là nghiệm của phương trình
c/ \(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=2y-2x\left(1\right)\\x^3+1=2y\left(2\right)\end{matrix}\right.\)
(1)\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
\(\Leftrightarrow x=y\)(vì x^2 +xy+y^2 +2>0)
thay x=y vào (2) ta được:
\(x^3-2x+1=0\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\sqrt{5}-1}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
vậy...
