Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4

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

\(\Leftrightarrow2x^2+8x-2x^2-x+3=0\)

\(\Leftrightarrow7x=-3\Leftrightarrow x=-\dfrac{3}{7}\)

b) \(x^2-2x-3=0\)

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

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

\(a,\Leftrightarrow2x^2+8x-2x^2-x+3=0\\ \Leftrightarrow7x=-3\\ \Leftrightarrow x=-\dfrac{3}{7}\\ b,x^2-2x-3=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

a: Ta có: \(2x\left(x+4\right)-\left(x-1\right)\cdot\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+8x-2x^2-3x+2x+3=0\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

b: ta có: \(x^2-2x-3=0\)

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

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

\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)

  Vậy ...

b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)

  Vậy ...

a: Ta có: \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x=5\)

hay x=5

b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)

\(\Leftrightarrow6x^2+8x-6x^2=16\)

\(\Leftrightarrow8x=16\)

hay x=2

a: Ta có: \(2^{x-1}=32\)

\(\Leftrightarrow x-1=5\)

hay x=6

b: Ta có: \(3^{2x+1}=81\)

\(\Leftrightarrow2x+1=4\)

\(\Leftrightarrow2x=3\)

hay \(x=\dfrac{3}{2}\)

c: Ta có: \(2^x-26=6\)

\(\Leftrightarrow2^x=32\)

hay x=5

d: Ta có: \(27\cdot3^x=243\)

\(\Leftrightarrow3^x=9\)

hay x=2

a) 70 - 5(x - 3 ) = 45

5( x - 3 ) = 70 - 45 = 25

x - 3 = 25 : 5 = 5

x = 5 + 3 = 8

b) (2x - 1 )⁴ = 3 . 6² - 27

(2x - 1 )⁴ = 3 . 36 - 27

(2x - 1 )⁴ = 81

 Ta thấy 81  = 3⁴ vậy suy ra (2x - 1)⁴ = 3⁴

 Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2

Thử lại : 2 . 2 - 1 = 4 - 1 = 3

Vậy x = 2

c) 3x³ +  43 = 10² - 33

3x³ + 43 = 100 - 33 = 67

3x³ = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)

`a)4x(x+1)+(3-2x)(3+2x)=15`

`<=>4x^2+4x+9-4x^2=15`

`<=>4x=6`

`<=>x=3/2`

Vậy `S={3/2}`

`b)3x(x-20012)-x+20012=0`

`<=>3x(x-20012)-(x-20012)=0`

`<=>(x-20012)(3x-1)=0`

`<=>` $\left[\begin{matrix} x=20012\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S={1/3;20012}`

a) 4x(x + 1) + (3 – 2x)(3 + 2x) = 15

⇔4x² + 4x + (9 – 4x²) = 15

⇔ 4x² + 4x + 9 – 4x² = 15

⇔4x = 15 – 9

⇔x=1,5

b)3x(x – 2001²) – x + 2001² = 0

⇔3x(x – 2001²) – (x – 2001²) = 0

⇔(x – 2001²)(3x – 1) = 0

⇔x – 2001² = 0 hay 3x – 1 = 0

⇔x = 2001² hoặc x = \(\dfrac{1}{2}\)

a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)

\(\Rightarrow2x^2+x-6-4x^2+22x-10=-16\)

\(\Rightarrow2x^2-23x=0\Rightarrow x\left(2x-23\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)

b) \(7x^2-7=x^2-2x+1\)

\(\Rightarrow7\left(x^2-1\right)-\left(x^2-2x+1\right)=0\)

\(\Rightarrow7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)\left(7x+7-x+1\right)=0\Rightarrow2\left(x-1\right)\left(3x+4\right)=0\)

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

a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)

 \(2x^2+x-6-4x^2+22x-10=-16\)

 \(-2x^2+23x-16=-16\)

\(23x-2x^2=0\)

\(x\left(23-2x\right)=0\)

⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)

b) \(7x^2-7=x^2-2x+1\)

\(7\left(x^2-1\right)=\left(x-1\right)^2\)

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

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

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

\(\left(x-1\right)\left(6x+8\right)=0\)

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