kho..............wa...................troi................thi......................ret.....................ai..............tich...............ung.....................ho....................minh..................voi................ret............wa

ặc toán lớp 8 thì khó quá trời

b) x(x-4) - 2x+8 = 0
    x(x-4) - 2(x-4) = 0
    (x-2) (x-4) = 0
TH1: x-2=0              TH2: x-4=0
            x=2                          x=4
Vậy x\(\in\){2;4}

\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)

b) x(x-4)-2x+8=0

x(x-4)-2(x-4)=0

(x-4)(x-2)=0

th1: x-4=0

x=4

th2: x-2=0

x=2

Vậy x thuộc tập hợp 4;-2

Học sinh tự nhẩm và ghi như sau:

8 x 3 = 24

8 x 2 = 16

8 x 4 = 32

8 x 1 = 8

8 x 5 = 40

8 x 6 = 48

8 x 7 = 56

0 x 8 = 0

8 x 8 = 64

8 x 10 = 80

8 x 9 = 72

8 x 0 = 0

1.24

2.16

3.32

4.8

5.40

6.48

7.56

8.0

9.64

10.80

11.72

12.0

c) =>x mũ 2 -9<0 hoăc x mũ 2 +1 <0

Mà x mũ 2+1>0 => x mũ 2+1 ko thể <0 

=>x mũ 2 -9< 0

=>x mũ 2 <9

=>x<3 hoặc x> -3 

Mk làm nhầm bạn sửa dấu > và < thành dấu bằng là đc nhé sorry

a) -3(x-4)+5(x-1)=-7

=>-3x+12+5x-5=-7

=>2x+7=-7

=>2x=-14=>x=-7

b) -4./x-8/+12=0

=>/x-8/=3

=>x-8=3 hoặc -3

(tự tính) 

a: \(3\left(x-3\right)-6x=0\)

=>\(3x-9-6x=0\)

=>-3x-9=0

=>3x+9=0

=>3x=-9

=>\(x=-\dfrac{9}{3}=-3\)

b: Đề thiếu vế phải rồi bạn

c: \(2\left(x-3\right)+3x=9\)

=>2x-6+3x=9

=>5x-6=9

=>5x=6+9=15

=>x=15/5=3

d: \(x\left(x-11\right)+2\left(x-11\right)=0\)

=>\(\left(x-11\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)

e: \(x\left(x+2\right)+8=x^2\)

=>\(x^2+2x+8=x^2\)

=>2x+8=0

=>2x=-8

=>x=-8/2=-4

f: \(8\left(x+1\right)+2x=-2\)

=>\(8x+8+2x=-2\)

=>10x=-2-8=-10

=>\(x=-\dfrac{10}{10}=-1\)

g: 12-3(x+2)=0

=>3(x+2)=12

=>x+2=12/3=4

=>x=4-2=2

\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)