a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

đỉ mẹ, đỉ má, cái lồn, con cặc.

Câu a:

(x - 5)^2 - 2x - 2^4 = 2x

(x - 5)(x - 5) - 2x - 16 - 2x = 0

x^2 - 5x - 5x + 25 - 2x - 16 - 2x = 0

x^2 - (5x + 5x +2x + 2x) + (25 - 16) = 0

x^2 - 14x + 9 = 0

(x^2 - 7x) - (7x - 49) - 40 = 0

x(x - 7) - 7(x - 7) - 40 = 0

(x - 7)(x - 7) - 40 = 0

(x - 7)^2 = 40

x - 7 = \(\sqrt{40}\) hoặc x - 7 = -\(\sqrt{40}\)

x - 7 = - \(\sqrt{40}\)

x = 7 - \(\sqrt{40}\)

x - 7 = \(\sqrt{40}\)

x = 7+ \(\sqrt{40}\)

Vậy x ∈ {7 - \(\sqrt{40}\) ; 7+ \(\sqrt{40}\))

Câu b:

3^(x -1) - 7^2 = 2^5 + 0^3

3^(x -1) - 49 = 32 + 0

3^(x - 1) - 49 = 32

3^(x -1) = 32 + 49

3^(x -1) = 81

3^(x-1) = 3^4

x - 1 = 4

x = 4 + 1

x = 5

Vậy x = 5

1000 g đó chọn mình đi sau mình tich cho bạn

a)

•  \(A=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{59}.3\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

• \(A=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{58}.7\)

\(=7\left(2+2^4+2^{58}\right)⋮7\)

• \(A=2+2^2+2^3+...+2^{60}\)​

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=2.15+2^5.15+...+2^{57}.15\)

\(=15\left(2+2^5+2^{57}\right)⋮15\)

b) \(B=1+5+5^2+5^3+...+5^{96}+5^{97}+5^{98}\)

\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)

\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+..+5^{96}\left(1+5+5^2\right)\)

\(=31+5^3.31+...+5^{96}.31\)

\(=31\left(1+5^3+...+5^{96}\right)⋮31\)

Ta có B = 3 + 3² + 3³ + ... + 3²⁰¹⁴ + 3²⁰¹⁵

=> 3B = 3² + 3³ + 3⁴ + .... + 3²⁰¹⁵ + 3²⁰¹⁶

Lấy 3B trừ B theo vế ta có 

3B - B = (3² + 3³ + 3⁴ + .... + 3²⁰¹⁵ + 3²⁰¹⁶) - (3 + 3² + 3³ + ... + 3²⁰¹⁴ + 3²⁰¹⁵)

   2B   = 3²⁰¹⁶ - 3

Khi đó 2B + 3 = 3^x

<=>  3²⁰¹⁶ - 3 + 3 = 3^x

=> 3²⁰¹⁶ = 3^x

=> x = 2016 

Vậy x = 2016

Bg

Ta có: B = 3 + 3² + 3³ +...+ 3²⁰¹⁴ + 3²⁰¹⁵ 

=> 3B = 3.(3 + 3² + 3³ +...+ 3²⁰¹⁴ + 3²⁰¹⁵)

=> 3B = 3.3 + 3.3² + 3.3³ +...+ 3.3²⁰¹⁴ + 3.3²⁰¹⁵

=> 3B = 3² + 3³ + 3⁴ +...+ 3²⁰¹⁵ + 3²⁰¹⁶ 

=> 3B - B = (3² + 3³ + 3⁴ +...+ 3²⁰¹⁵ + 3²⁰¹⁶) - (3 + 3² + 3³ +...+ 3²⁰¹⁴ + 3²⁰¹⁵)

=> 2B = 3²⁰¹⁶ - 3

 2B + 3 = 3^x

=> 3²⁰¹⁶ - 3 + 3 = 3^x

=> 3²⁰¹⁶ = 3^x

=> x = 2016