a, \(x\left(x-\dfrac{1}{2}\right)< 0\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-\dfrac{1}{2}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-\dfrac{1}{2}>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\dfrac{1}{2}< x< 0\)

Vậy...

b, \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|=0\)

Mà \(\left|\dfrac{1}{4}-x\right|+\left|x-y+z\right|+\left|\dfrac{2}{3}+y\right|\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}-x=0\\x-y+z=0\\\dfrac{2}{3}+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\z=\dfrac{-11}{12}\\y=\dfrac{-2}{3}\end{matrix}\right.\)

Vậy...

+) Xét \(x\ge0\) có:

\(x+x-1=7\Rightarrow x=4\) ( t/m )
+) Xét \(-1\le x< 0\) có:
\(-x+x-1=7\Rightarrow-1=7\) ( vô lí )
+) Xét x < -1 có:
\(-x-x+1=7\Rightarrow x=-3\) ( t/m )

Vậy...

\(\left|x\right|+\left|x-1\right|=7\)

+) Xét \(x\ge0\) có:
\(x+x-1=7\Rightarrow x=4\) ( t/m )

+) Xét \(-1\le x< 0\) có:

\(-x+x-1=7\Rightarrow-1=7\) ( vô lí )

+) Xét \(x< -1\) có:
\(-x-x-1=7\Rightarrow x=-4\) ( t/m )

Vậy x = 4 hoặc x = -4

+) Xét \(2x+3y-1=0\) có:

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

+) Xét \(2x+3y-1\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

\(\Rightarrow y=3\)

Vậy...

\(27y^3-9y^2+y-\dfrac{1}{27}\)

\(=\left(3y-\dfrac{1}{3}\right)^3\)

Ta có: \(-3< 0\)

\(x^2+1\ge1>0\forall x\)

\(\Rightarrow A=\dfrac{-3}{x^2+1}< 0\forall x\)

\(\Rightarrowđpcm\)

a, \(A=2x^2+4x+5=2x^2+4x+2+3\)

\(=2\left(x+1\right)^2+3>0\)

\(\Rightarrowđpcm\)

b, \(B=-3x^2+6x-7=-3x^2+6x-3-4\)

\(=-3\left(x-1\right)^2-4< 0\)

\(\Rightarrowđpcm\)

mk chỉ biết copy khi kiểm tra thôi

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