a: Ta có: \(x^2-2x=0\)

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

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

b: Ta có: \(\left(x-1\right)\cdot x-2\left(1-x\right)=0\)

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

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

c: Ta có: \(x^3+2x^2+x=0\)

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

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

d: Ta có: \(x^3-3x^2=0\)

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

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

a) x² - 5x - 6 = 0

=> x² - 2x - 3x - 6 = 0

=> (x² - 2x) + (-3x - 6) = 0

=> x(x - 2) - 3 (x - 2) = 0

=> (x - 2) (x - 3) = 0

=> x - 2 = 0 => x = 2

     x - 3 = 0 => x = 3

còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546

Bạn giải giúp mình mấy câu kia được k=)))

a.\(x^3-x=0 \)

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

x=0 hay x²-1=0

x=0 hay x²=1

x=0 hay x=1

Vậy x=0 hay x=1

b.\(x^3+1=0\)

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

\(x=0 hay x^2+1=0\)

\(x=0 hay x^2=-1\)(vô lí vì x²≥0)

Vậy x=0

c.\(x^2-4x=0\)

\(x(x-4)=0\)

x=0 hay x-4=0

x=0 hay x=4

Vậy x=0 hay x=4

d.\(x(x-1)-2(1-x)=0\)

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

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

x-1=0 hay x+2=0

x=1 hay x=-2

Vậy x=1 hay x=-2

e.\(2x(x-2)-(2-x)^2=0\)

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

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

\((x-2)(3x-2)=0\)

x-2=0 hay 3x-2=0

x=2 hay 3x=2

x=2 hay x=2/3

Vậy x=2 hay x=2/3

f.\(4x(x+1)=8(x+1)\)

\(4x(x+1)-8(x+1)=0\)

\(4(x+1)(x-2)=0\)

_{4(x+1)=0 hay x-2=0}

_{x+1=0 hay x=2}

_{x=-1 hay x=2}

_{Vậy x=-1 hay x=2}

_{g.\(5x(x-2)-x+2=0\)}

\(5x(x-2)-(x-2)=0\)

\((x-2)(5x-1)=0\)

x-2=0 hay 5x-1=0

x=2 hay 5x=1

x=2 hay x=1/5

Vậy x=2 hay x=1/5

h.\((x+1)=(x+1)^2\)

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

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

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

x+1= 0 hay -x=0

x=-1 hay x=0

Vậy x=-1 hay x=0

a; (\(x\) - 2)².(\(x+1\)).(\(x\) - 4) < 0

    (\(x-2\))² ≥ 0 ∀\(x\); \(x+1\) = 0 ⇒ \(x=-1\); \(x-4\) = 0 ⇒ \(x=4\)

Lập bảng ta có:

Theo bảng trên ta có: -1 < \(x\) < 4

Vậy \(-1< x< 4\)

b; [\(x^2\).(\(x-3\)):(\(x-9\))] < 0

    \(x-3=0\)⇒ \(x=3\); \(x-9\) = 0 ⇒ \(x=9\)

    Lập bảng ta có:

Theo bảng trên ta có:     3 < \(x\) < 9

Vậy 3 < \(x\) < 9

a.x(y+3)=3

=> x(y+3) ∈Ư(3)={-3;-1;1;3}

ta có bảng sau

vậy x=-3 thì y=-4

x=-1 thì y=-6

x=1 thì y=0

x=3 thì y=-2

c.x+3⋮ x+1

=> (x+3)-(x+1)⋮(x+1)

=> (x+3-x-1)⋮(x+1)

=> 2⋮(x+1)

=> (x+1) ∈ Ư(2)={-2;-1;1;2}

=> x∈{-3;-2;0;1}

vậy x ∈{-3;-2;0;1}

b,d tương tự

a.(x-2)(x+3)>0

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

=> x>2

vậy x>2

b.(x-2)(x-1)>0

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

=> x>2

vậy x>2

c.(x-2)(x²+1)>0

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

vậy x>2

d.(x-1)(x+2)>0

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

=> x>1

vậy x>1

Còn câu này nx bn ạ:

x^2.(x+2)<0

Tìm x

Giúp mk nhanh nha, mk cần gấp

a.x²+x=0

 => x(x+1)=0

 => x thuộc {0;-1}

b.(x-1)^x+2=(x-1)^x+4

 => [x²-2x](x-1)^x+2=0

 =>(x-2)x(x-1)^x+2 =0

=> x thuộc {0;1;2}

c.Ta có (2x-1)³=-8

       =>2x-1=-2

       => 2x=-1

       => x=-1/2

a, \(x^3-x^2=4x^2-8x+4\)

\(\Rightarrow x^3-x^2-4x^2+8x-4=0\)

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

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

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

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

b, \(\left(x-1\right)\left(x^2+x+1\right)=7\)

\(\Rightarrow x^3-1=7\Rightarrow x^3=2^3\Rightarrow x=2\)

c, \(2\left(x+5\right)-x^2-5x=0\)

\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

d, \(x^2-3x=-2\)

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

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

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

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

Chúc bạn học tốt!!!