1.

Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)

Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)

\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)

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

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

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

Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):

\(-x^2+x+1=-x^2+3x\)

\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)

Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\): 

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)

Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)

Bài 2:

x^3+6x^2+12x+m chia hết cho x+2

=>x^3+2x^2+4x^2+8x+4x+8+m-8 chia hết cho x+2

=>m-8=0

=>m=8

a) \(x^3-16x=0\)

 ⇔\(x\left(x^2-16\right)=0\)

 ⇒\(x=0\) hoặc \(x^2-16=0\)

\(TH_1:x=0\)

\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)

             Vậy \(x\in\left\{0;\pm4\right\}\)

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

⇒ \(2x+1=x-1\)

⇒ \(2x+2=x\)

⇒ \(2\left(x+1\right)=x\) ⇒ x = -2 

        Vậy x = -2

f'(x)=(m-1)*x^2+2

Đặt f'(x)=0

=>\(x^2=\dfrac{-2}{m-1}\)

=>\(x=\pm\sqrt{-\dfrac{2}{m-1}}\)

Để phương trình có nghiệm dương thì m-1<0

=>m<1

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a,x(2x-1)-(x-1)^2-x^2=0

<=>x(2x-1-x)-(x-1)^2=0

<=>x(x-1)-(x-1)^2=0

<=>(x-x+1)(x-1)=0

<=>x-1=0

<=>x=1

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

<=>x^3+6x^2+12x+8-x^3-6x^2=4

<=>12x+8=4

<=>x=-1/3

tick mik nha

`a)x(2x-1)-(x-1)^2-x^2=0`

`<=>2x^2-x-x^2+2x-1-x^2=0`

`<=>x-1=0`

`<=>x=1`

Vậy `x=1.`

`b)(x+2)^3-x^3-6x^2=4`

`<=>x^3+6x^2+12x+8-x^3-6x^2=4`

`<=>12x+8=4`

`<=>12x=-4`

`<=>x=-1/3`

Vậy `x=-1/3.`

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

\(\Leftrightarrow2x^2-x-x^2+2x-1-x^2=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(\left(x+2\right)^3-x^3-6x^2=4\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=4\)

\(\Leftrightarrow12x=-4\)

hay \(x=-\dfrac{1}{3}\)

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

a, (x+3)² - ( 2x + 1 ).( x+3)=0              b,     x³-12x²+36x =0

=> (x+3).(x+3-2x-1)                             => x(x²-12x+36) = 0

=>(x+3).(-x+2)                                     => x(x-6)² = 0

=> x+3=0  <=> x=-3                            => x=0        <=> x=0

     -x+2=0 <=> x=-2                                 x-6= 0    <=> x=6