a) Ta có: \(3x-1=0\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) Ta có: \(5x-2=x+4\)

\(\Leftrightarrow5x-x=4+2\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

\(\left(x.x+1\right)\left(x-1\right)\left(x.x.x+1\right)=0\)

TH1 : \(x.x+1=0\Leftrightarrow x^2+1=0\Rightarrow x^2=-1\) ( loại )

TH2 : \(x-1=0\Rightarrow x=1\) ( chọn )

TH3 : \(x.x.x+1=0\Leftrightarrow x^3+1=0\Rightarrow x^3=-1\Rightarrow x=-1\) ( chọn )

Vậy có 2 giá trị của x thỏa mãn đề bài

Ta có B = x x + x + 1 x − 1 x + x + 1 + x + 3 x − 1 . x − 1 2 x + x − 1

= x x − 1 + x + 3 x − 1 . x − 1 x + 1 2 x − 1 x + 1 = 2 x + 3 x − 1 . x − 1 2 x − 1 = 2 x + 3 2 x − 1    

Vì x ≥ 0  nên  2 x + 3 > 0 , do đó B<0 khi  2 x − 1 < 0 ⇔ x < 1 4 .

Mà  x ≥ 0 ;   x ≠ 1  và  x ≠ 1 4  nên ta được kết quả  0 ≤ x < 1 4 .

a) \(\left(x+4\right)\left(x\cdot x+1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x^2+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x\notin R\end{matrix}\right.\)

\(\Rightarrow x=-4\)

Vậy \(x=-4\)

b) \(\left(\left|x\right|+2\right)\left(x\cdot x-1\right)=0\)

\(\Rightarrow\left(\left|x\right|+2\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x\right|+2=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=1\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy \(x_1=-1;x_2=1\)

Tìm x \(\in\)Z

a) (x+4)(x.x+1)=0

(x+4)(x²+1)=0

<=>x+4 và x²+1=0

Nếu x+4=0=>x=-4

Nếu x²+1=0=>x²=-1(ko có giá tri)

Vậy...

b)(|x|+2)(x.x-1)=0

(|x|+2)(x²-1)=0

<=>|x|+2 và x²-1=0

+|x|+2=0=>|x|=-2(không tồn tại)

+x²-1=0=>x²=1=>x={1;-1}

Vậy....

a)\(\left(x+4\right)\left(x\cdot x+1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow x=-4\)

Vậy \(x=-4\).

b)\(\left(\left|x\right|+2\right)\left(x\cdot x-1\right)=0\)

\(\Rightarrow\left(\left|x\right|+2\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x\right|+2=0\\x^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=1\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\).