a) \(2\left(x+5\right)-4x=1\)
\(\Leftrightarrow-2x+9=0\)
\(\Rightarrow x=\dfrac{9}{2}\)
b) \(\left(x-4\right)\left(5x-2\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5x-2\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5x-2+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\5x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-1}{5}\end{matrix}\right.\)
Vậy..
c) \(x^4-8x^2-9=0\)
\(\Leftrightarrow\left(x^2\right)^2-2x^24+16-25=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-5^2=0\)
\(\Leftrightarrow\left(x^2-4-5\right)\left(x^2-4+5\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(x^2+1\right)=0\)
Vì x2 + 1 > 0
\(\Rightarrow x^2-9=0\)
\(\Rightarrow x=\pm3\)
a, phần này bạn tự làm nhé
b, \(\left(x-4\right)\left(5x-2\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow5x^2-22x+8-12+3x=0\)
\(\Leftrightarrow5x^2-19x-4=0\)
\(\Leftrightarrow5x^2-20x+x-4=0\)
\(\Leftrightarrow5x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(5x+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=4\end{matrix}\right.\)
Vậy...
c, \(x^4-8x^2-9=0\)
\(\Leftrightarrow x^4-9x^2+x^2-9=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)+\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
