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\(x^2-x-\frac{3}{4}=0\)
\(\Rightarrow\frac{x^2-x}{1}-\frac{3}{4}=0\)
\(\Rightarrow\frac{\left(x^2-x\right).4}{4}-\frac{3}{4}=0\)
\(\Rightarrow\frac{4x^2-4x-3}{4}=0\)
\(\Rightarrow4x^2-4x-3=0\)
\(\Rightarrow4x^2-6x+2x-3=0\)
\(\Rightarrow2x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}}\)
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Mà 1-2y lẻ nên:
\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)
b tương tự.
c) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)
d tương tự
\(x^2-x-\frac{3}{4}=0\)
\(\Rightarrow x^2+\frac{1}{2}x-\frac{3}{2}x-\frac{3}{4}=0\)
\(\Rightarrow x\left(x+\frac{1}{2}\right)-\frac{3}{2}\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\left(x-\frac{3}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
Chúc bạn học tốt.
nhưng mà mình chỉ biết thế thôi thiếu gì thì bổ sung, còn làm dc thì làm luôn nha đung mình k
1) |x|=x+2
=> \(\left[{}\begin{matrix}x=x+2\\x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(voli\right)\\2x=-2\Rightarrow x=-1\end{matrix}\right.\)
vậy x=-1
c;b tương tự
2) \(\left|x-\dfrac{3}{2}\right|=\left|\dfrac{5}{2}-x\right|\)
=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{5}{2}-x\\x-\dfrac{3}{2}=x-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\Rightarrow x=2\\0=-1\left(voli\right)\end{matrix}\right.\)
vậy x=2
