Dễ mà,e cứ chia 2 TH là đc

Vd:<0 thì chia ra x+2>0 hoac x<0 và nguoc lai roi tìm x

9 x 1 = 9

9 x 5 = 45

9 x 4 = 32

9 x 10 = 90

9 x 2 = 18

9 x 7 = 63

9 x 8 = 72

9 x 0 = 0

9 x 3 = 27

9 x 9 = 81

9 x 6 = 54

0 x 9 = 0

= 9                            = 45                                      = 36

dễ mà bạn gì ơi

9 x 4 = 36

9 x 2 = 18

9 x 5 = 40

9 x 10 = 90

9 x 1 = 9

9 x 7 = 63

9 x 8 = 72

0 x 9 = 0

9 x 3 = 27

9 x 6 = 54

9 x 9 = 81

9 x 0 = 0

a) 9 x 1=9                   9 x 4=36                   9 x 7=63

b) 9 x 2=18                 9 x 5=45                   9 x 8=72

c) 9 x 3=27                 9 x 6=54                   9 x 981

d) 9 x 10=90               9 x 0=0                     0 x 9=0

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x+3}}\)(\(x\ge0,x\ne9\))

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\sqrt{x}-2\left(x\ge0,x\ne9\right)\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)

mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)

a,\(\dfrac{3-\sqrt{x}}{x-9}\)

=\(-\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

=\(-\dfrac{1}{3+\sqrt{x}}\)

6 x 1 = 6              6 x 9 = 54

6 x 2 = 12              6 x 8 = 48

6 x 3 = 18              6 x 7 = 42

6 x 4 = 24              0 x 6 = 0

6 x 5 = 30              6 x 0 = 0

6 x 6 = 36              6 x 10 = 60

6 x 4 =24

6 x 1 =6

6 x 9 =54

6 x 10 =60

6 x 6 =36

6 x 3 =18

6 x 2 =12

0 x 6 =0

6 x 8 =48

6 x 5 =30

6 x 7 =42

6 x 0 =0

a,

đoạn 9x-6-> 2x-6=0

=> x=3

b,6x^2+13x+5=6x^2-20x+6

33x=1

=>x=1/33

a) (x+1)(x+9)=(x+3)(x+5) 

<=>x^2+10x+9=x^2+8x+15

<=>x^2+10x+9-x^2-8x-15=0

<=>9x-6=0 phải là 2x - 6

<=>9x=6

<=>x=6/9=2/3 => S= 2/3

d) (3x+5)(2x+1)=(6x-2)(x-3)

<=>6x^2+13x+5=6x^2-16x+6 phải là 6x^2 - 20x + 6

<=>6x^2+13x+5-6x^2+16x-6=0

<=>29x-1=0

<=>29x=1

<=>x=1/29

a) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)

\(\Leftrightarrow x^2+10x+9-x^2-8x-15=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\)

Vây tập nghiệm của phương trình là \(S=\left\{3\right\}\)

d) \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow x=\frac{1}{33}\)

Vây tập nghiệm của phương trình là \(S=\left\{\frac{1}{33}\right\}\)