a) 

\(\left(3^2-2^3\right)x+3^2\cdot2^2=4^2\cdot3\\ \left(9-8\right)x+36=48\\ x=48-36\\ x=12\)

b) 

\(\dfrac{x-2}{-4}=\dfrac{-9}{x-2}\left(x\ne2\right)\\\left(x-2\right)^2 =-4\cdot-9\\ \left(x-2\right)^2=36\\ \left(x-2\right)^2=6^2 \)

TH1: x - 2 = 6 

x = 8

TH2: x - 2 = -6

x = -4

c) 

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\\ \left(x-1\right)^2+9=25\cdot1\\ \left(x-1\right)^2=25-9=16\\ \left(x-1\right)^2=4^2\)

TH1:  x - 1 = 4

x = 5

TH2: x - 1 = -4

x = -3

d) x⁵-x³=0

⇔ x³(x²-1)=0

⇔ x³(x-1)(x+1)=0

TH1: x=0

TH2: x-1=0 ⇔ x =1

TH3: x + 1=0 ⇔ x =-1

c) x-5 / 3 = -12 /5-x

⇔ x-5 / 3  = 12 / x-5 

⇔ (x-5)²= 36

⇔ (x-5)²= 6²

^{TH1: x -5 =6} 

⇔ x = 11

TH2: x- 5 = -6

⇔ x = -1

f)  ⇔ (2x-1)(2x-2)=0

TH1: 2x-1 = 0 

⇔ x = 1/2

TH2: 2x-2 = 0

⇔ x=1

g) bài toán có quy luật

⇔ \(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2023}{2024}\)   (nhân 2 cả tử và mẫu)

⇔\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2023}{4048}\)   ( chia hai vế cho 2 )

⇔\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2023}{4048}\)

⇔\(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{2023}{4048}\) 

⇔\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\)

⇔\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2023}{4048}\)    ( rút gọn ta đc)

⇔\(\dfrac{\left(x+1\right)-2}{2.\left(x+1\right)}=\dfrac{2023}{4048}\) ( quy đồng)

⇔ \(\dfrac{x-1}{x+1}=\dfrac{2023}{2024}\)    ( nhân 2 vế cho 2 )

⇔ 2024.(x-1)= 2023. ( x+1)

⇔ 2024x -2024 = 2023x + 2023

⇔ 2024x - 2023x= 2023+2024

⇔ x = 4047

tick cho tui nha! 

Quãng đường Nam đi được trong 3 giờ đầu là:

\(12\cdot3=36\left(km\right)\)

Quãng đường Nam đi được trong 2 giờ sau là:

\(11\cdot2=22\left(km\right)\)

Quãng đường Nam đi được trong 5 giờ là:

\(36+22=58\left(km\right)\)

Vậy...

Tổng quãng đường mà bạn Nam đi được trong 3 giờ đầu là:

\(3\times12=36\left(km\right)\)

Tổng quãng đường mà bạn Nam đi được trong 2 giờ sau là:

\(2\times11=22\left(km\right)\)

Tổng quãng đường mà bạn Nam đi được trong 5 giờ là:

\(36+22=58\left(km\right)\)

Đáp số : \(58km\)

khó thế ?

right

khó thiệt

15583 là số nguyên tố bạn nhé

ê

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\\ =\left[\left(\dfrac{2}{3}\right)^2\right]^4:\left(\dfrac{4}{9}\right)^3\\ =\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3\\ =\dfrac{4}{9}\)

\(27^3:3^2\\ =\left(3^3\right)^3:3^2\\ =3^9:3^2\\ =3^7\\ =2187\)

\(\left(-\dfrac{3}{5}\right)^4:\left(\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{3}{5}:\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{6}{25}\right)^4\)

\(\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{9}{25}\right)^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left[\left(\dfrac{3}{5}\right)^2\right]^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{3}{5}\right)^{10}\\ =\left(\dfrac{3}{5}\right)^2\\ =\dfrac{9}{25}\)

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3=\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3=\dfrac{4}{9}\)

\(27^3:3^2=3^9:3^2=3^7=2187\)

a) Ta có: 

\(64^8=\left(2^6\right)^8=2^{6\cdot8}=2^{48}\)

\(16^{12}=\left(2^4\right)^{12}=2^{4\cdot12}=2^{48}\)

\(\Rightarrow64^8=16^{12}\)

b) Ta có:

\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{4\cdot10}=\left(\dfrac{1}{2}\right)^{40}\)

Mà: 50 > 40 => `(1/2)^50<(1/2)^40` 

c) Ta có: 

\(\left(\dfrac{9}{16}\right)^{100}=\left[\left(\dfrac{3}{4}\right)^2\right]^{100}=\left(\dfrac{3}{4}\right)^{200}\)

Mà: `3/4>2/3=>(3/4)^200>(2/3)^200`

\(^{^{ }}\)a,64^8=16^12

b,(1/16)^10<(1/2)^50

c,(2/3)^200>(9/16)^100

CỦA BẠN ĐÂY NẾU SAI THÌ CHO MÌNH XIN LỖI NHÉ

a; 25 x 5³ x \(\dfrac{1}{625}\) x 5²

   = 5² x 5³ x \(\dfrac{1}{5^4}\) x 5²

= 5⁵ x \(\dfrac{1}{5^4}\) x 5²

= 5 x 5²

= 5³

a) 

\(25\cdot5^3\cdot\dfrac{1}{625}\cdot5^2\\ =\left(5^2\cdot5^3\cdot5^2\right)\cdot\dfrac{1}{625}\\ =5^7\cdot\dfrac{1}{5^4}\\ =5^3\)

b) 

\(5^2\cdot3^5\cdot\left(\dfrac{3}{5}\right)^2\\ =5^2\cdot3^5\cdot\dfrac{3^2}{5^2}\\ =3^5\cdot3^2\\ =3^7\)

c) 

\(\left(-\dfrac{1}{7}\right)^4\cdot49^2\\ =\dfrac{\left(-1\right)^4}{7^4}\cdot\left(7^2\right)^2\\ =\dfrac{1}{7^4}\cdot7^4\\ =1\)

d) 

\(\left(\dfrac{1}{16}\right)^2:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{8}\right)^3\\ =\left[\left(\dfrac{1}{2}\right)^4\right]^2:\left(\dfrac{1}{2}\right)^4\cdot\left[\left(-\dfrac{1}{2}\right)^3\right]^3\\ =\left(\dfrac{1}{2}\right)^8:\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot\left(-\dfrac{1}{2}\right)^9\\ =\left(\dfrac{1}{2}\right)^4\cdot-\left(\dfrac{1}{2}\right)^9\\ =-\left(\dfrac{1}{2}\right)^{13}\)

a; \(x\) : (- \(\dfrac{1}{3}\))³ = \(\dfrac{1}{9}\)

    \(x\) : (\(-\dfrac{1}{27}\)) = \(\dfrac{1}{9}\)

    \(x\)               = \(\dfrac{1}{9}\) x (- \(\dfrac{1}{27}\))

    \(x\)               = - \(\dfrac{1}{243}\)

    Vậy  \(x\)    =  - \(\dfrac{1}{243}\)

b; (\(\dfrac{4}{5}\))⁵ x \(x\) = (\(\dfrac{4}{5}\))⁷

               \(x\) = (\(\dfrac{4}{5}\))⁷ : (\(\dfrac{4}{5}\))⁵

               \(x\) = \(\dfrac{4^7}{5^7}\) : \(\dfrac{4^5}{5^5}\)

               \(x\) = \(\dfrac{4^7}{5^7}\) x \(\dfrac{5^5}{4^5}\)

               \(x\) = \(\dfrac{4^2}{5^2}\)

               \(x\) = \(\dfrac{16}{25}\)

Vậy  \(x\) = \(\dfrac{16}{25}\)