how do you spell your name 

a (x + 2) - x(x + 3) = 2

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

x + x(x + 3) = 0

x(1 + x + 3) = 0

x(x + 4) = 0

x = 0 hoặc x + 4 = 0

*) x + 4 = 0

x = -4

Vậy x = -4; x = 0

b) (x + 2)(x - 2) - (x + 1)² = 7

x² - 4 - x² - 2x - 1 = 7

-2x - 5 = 7

-2x = 7 + 5

-2x = 12

x = 12 : (-2)

x = -6

c) 6x² - (2x + 1)(3x - 2) = 1

6x² - 6x² + 4x - 3x + 2 = 1

x + 2 = 1

x = 1 - 2

x = -1

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

x² + 3x + 2x + 6 - x² - x + 2x + 2 = 2

6x + 8 = 2

6x = 2 - 8

6x = -6

x = -6 : 6

x = -1

e) 6(x - 1)(x + 1) - (2x - 1)(3x + 2) + 3 = 0

6x² - 6 - 6x² - 4x + 3x + 2 + 3 = 0

-x - 1 = 0

x = -1

`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).

b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).

c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).

d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).

em cảm ơn ạ !

a: ta có: \(12⋮x+3\)

\(\Leftrightarrow x+3\in\left\{3;4;6;12\right\}\)

hay \(x\in\left\{0;1;3;9\right\}\)

b: Ta có: \(14⋮2x\)

\(\Leftrightarrow2x\in\left\{2;14\right\}\)

hay \(x\in\left\{1;7\right\}\)

3) Tìm số tự nhiên x, biết:

a) \(2^x=4\)

\(2^x=2^2\)

\(\Rightarrow x=2\)

___________

b) \(2^x=1\)

\(2^x=2^0\)

\(\Rightarrow x=0\)

___________

c) \(2^x=16\)

\(2^x=2^4\)

\(\Rightarrow x=4\)

___________

d) \(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

___________

e) \(5^x=125\)

\(5^x=5^3\)

\(\Rightarrow x=3\)

___________

f) \(8^x=64\)

\(8^x=8^2\)

\(\Rightarrow x=2\)

___________

h) \(3^{x+1}=3^2\)

\(\rightarrow x+1=2\)

\(x=2-1\)

\(x=1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

a: 2^x=4

=>2^x=2^2

=>x=2

b: 2^x=1

=>2^x=2^0

=>x=0

c: 2^x=16

=>2^x=2^4

=>x=4

d; 3^x=9

=>3^x=3^2

=>x=2

e: 5^x=125

=>5^x=5^3

=>x=3

f: 8^x=64

=>8^x=8^2

=>x=2

f: 3^x+1=3^2

=>x+1=2

=>x=1

\(a=0;1;2;3\) ở câu a

\(a=0;1;2;3;4;5;6;7\) ở câu b

\(a=0;1;2;3;4;5;6\) ở câu c

a) a = 3

b) b = 8

c) x = 1

d) ab = 23