mai mk học rồi , giúp nha, I x I là giá trị tuyệt đối của x !!!

b) Giải:
Ta có: \(4x+3⋮x-2\)

\(\Rightarrow4x-8+11⋮x-2\)

\(\Rightarrow4\left(x-2\right)+11⋮x-2\)

\(\Rightarrow11⋮x-2\)

\(\Rightarrow x-2\in\left\{1;-1;11;-11\right\}\)

\(\left[\begin{matrix}x-2=1\\x-2=-1\\x-2=11\\x-2=-11\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=13\\x=-9\end{matrix}\right.\)

Vậy \(x\in\left\{3;1;13;-9\right\}\)

b.Ta có:(4x+3)=4x-4.2+8+3

=4(x-2)+11

Để(4x+3)chia hết cho (x-2)

#11chia hết cho (x-2)(#là khi và chỉ khi nhế!)

#x-2€ Ư(11)={±1;±11}

#x€{3;1;13;-9}

Vậy x€{3;1;13;-9}

1 ) \(\left(x-4\right)^2-25=0\)

\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow-2\left(2x-4\right)=0\)

\(\Leftrightarrow x=2.\)

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)

5 ) \(x^3+x^2+x+1=0\)

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

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

6 ) \(x^3+x^2-x-1=0\)

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

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

7 ) \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=-1.\)

8 ) \(x^4-4x^3-19x^2+106x-120=0\)

\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)

\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)

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

9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)

\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)

Đặt \(x^2+6x-7=t\)

\(\Leftrightarrow t\left(t-9\right)+8=0\)

\(\Leftrightarrow t^2-9t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)

Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)

Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)

Vậy ........