Câu 1 : A

Câu 2 : D

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

\(b.\left(x-\frac{1}{2}\right)\left(2x+5\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-\frac{1}{2}=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{5}{2}\end{matrix}\right. \)

Câu \(b\) thấy hơi kì nên chắc đề như này.

\(c.x-2\left(\frac{2}{3}x-6\right)=0\\\Leftrightarrow x-\frac{4}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x+12=0\\\Leftrightarrow -\frac{1}{3}x=-12\\\Leftrightarrow x=36\)

\(d.x^2-2x=0\\\Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(e.\left(x^2-2x+1\right)-4=0\\ \Leftrightarrow\left(x-1\right)^2-4=0\\\Leftrightarrow \left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(f.x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)

\(g.4x^2+4x+1=0\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)=0\\\Leftrightarrow x^2+x+\frac{1}{4}=0\\\Leftrightarrow \left(x+\frac{1}{2}\right)^2=0\\\Leftrightarrow x+\frac{1}{2}=0\\ \Leftrightarrow x=-\frac{1}{2}\)

\(h.x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

\(i.2x^2+3x=0\\ \Leftrightarrow x\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\frac{3}{2}\end{matrix}\right.\)

\(\begin{array}{l} a)x\left( {{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 1\\ x = - 1 \end{array} \right.\\ b)\left( {x - \dfrac{1}{2}} \right)\left( {2x + 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{1}{2} = 0\\ 2x + 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{1}{2}\\ x = - \dfrac{5}{2} \end{array} \right.\\ c)\left( {x - 2} \right)\left( {\dfrac{2}{3}x - 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ \dfrac{2}{3}x - 6 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 9 \end{array} \right. \end{array}\)

a) \(x\left(x^2-1\right)=0\)

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

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

Vậy: x∈{-1;0;1}

d) \(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

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

Vậy: x∈{0;2}

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

\(\Leftrightarrow\left(x-1\right)^2-2^2=0\)

\(\Leftrightarrow\left(x-1-2\right)\left(x-1+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: x∈{3;-1}

f) \(x\left(2x-1\right)=0\)

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

Vậy: \(x\in\left\{0;\frac{1}{2}\right\}\)

g) \(4x^2+4x+1=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

hay \(x=\frac{-1}{2}\)

Vậy: \(x=\frac{-1}{2}\)

h) \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: x∈{2;3}

i) \(2x^2+3x=0\)

\(\Leftrightarrow x\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-3}{2}\right\}\)

(2\(x\) - 1).(2\(x\) - 5) < 0

Lập bảng ta có:

Theo bảng trên ta có: \(\dfrac{1}{2}\) < \(x\) < \(\dfrac{5}{2}\)

(3 - 2\(x\)).(\(x\) + 2) > 0

Lập bảng ta có:

Theo bảng trên ta có: - 2 < \(x\) < \(\dfrac{3}{2}\) 

(3\(x\) + 1).(5 - 2\(x\)) > 0

Lập bảng ta có:

Theo bảng trên ta có: - \(\dfrac{1}{3}\) < \(x\) < \(\dfrac{5}{2}\) 

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Khó quá,hay quá !!!

c(x-1)^2=4

x^2-2x+1=4

x^2-2x+1-4=0

x^2-2x-3=0

x^2-3x+x-3=0

x(x-3)+(x-3)=0

(x-3)(x+1)=0

\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)

d, x^3+2x^2-x-2=0

x^2(x+2)-(x+2)=0

(x+2)(x^2-1)=0

\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)

e, (3x+2)^2-(2x-1)^2=0

(3x+2-2x+1)(3x+2+2x-1)=0

(x+3)(5x-1)=0

x+3=0=>x=-3

5x-1=0=>5x=1=>x=1/5

a: =>(x-3)(x+1)=0

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

c: =>(x-5)(x+1)=0

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

h: =>x^2-4x+4-3=0

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

b: =>(x-4)(x-3)(x-1)>0

=>1<x<3 hoặc x>4

c: =>(2x-1)(x-1)(2x-3)<0

=>x<1/2 hoặc 1<x<3/2