Bài 6:
a: Thay m=-5 vào phương trình, ta được:
\(\left[4+\left(-5\right)^2\right]x-8x+2-\left(-5\right)=0\)
=>\(\left(4-25-8\right)x+7=0\)
=>-29x=-7
=>\(x=\dfrac{7}{29}\)
b: \(\left(4+m^2\right)x-8x+2-m=0\)
=>\(x\left(m^2+4-8\right)=m-2\)
=>\(x\left(m^2-4\right)=m-2\)
Để phương trình có nghiệm duy nhất thì \(m^2-4\ne0\)
=>\(m^2\ne4\)
=>\(m\notin\left\{2;-2\right\}\)
c: Thay x=1/4 vào phương trình, ta được:
\(\dfrac{1}{4}\left(m^2-4\right)=m-2\)
=>\(m^2-4=4\left(m-2\right)\)
=>(m+2)(m-2)-4(m-2)=0
=>(m-2)^2=0
=>m-2=0
=>m=2
Bài 4:
a: |x+9|=2
=>\(\left[{}\begin{matrix}x+9=2\\x+9=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2-9=-7\\x=-2-9=-11\end{matrix}\right.\)
b: \(\left|2x-3\right|=x-3\)
=>\(\left\{{}\begin{matrix}x-3>=0\\\left(2x-3\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-3-x+3\right)\left(2x-3+x-3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\x\cdot\left(3x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\left|2-x\right|=2x-1\)
=>\(\left|x-2\right|=2x-1\)
=>\(\left\{{}\begin{matrix}2x-1>=0\\\left(2x-1\right)^2=\left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x-1-x+2\right)\left(2x-1+x-2\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(x+1\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
d: \(\left|-2x\right|=x-3\)
=>|2x|=x-3
=>\(\left\{{}\begin{matrix}x-3>=0\\\left(2x\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-x+3\right)\left(2x+x-3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\\left(x+3\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
e: |3x-1|-x=2
=>|3x-1|=x+2
=>\(\left\{{}\begin{matrix}x+2>=0\\\left(3x-1\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x-1-x-2\right)\left(3x-1+x+2\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(2x-3\right)\left(4x+1\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{1}{4};\dfrac{3}{2}\right\}\)
f: |5x-4|=|x+1|
=>\(\left[{}\begin{matrix}5x-4=x+1\\5x-4=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=5\\6x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{3}{6}=\dfrac{1}{2}\end{matrix}\right.\)
