A = 7²⁰⁰⁸  = (7⁴)¹⁰⁰⁴ = (\(\overline{...1}\))¹⁰⁰⁴ = \(\overline{...1}\)

B = 2³⁴⁵⁶ = (2⁴)⁸⁶⁴ = \(\overline{...6}\)⁸⁶⁴ = \(\overline{...6}\)

C = 204²⁰⁸ = (204²)¹⁰⁴ = \(\overline{...6}\)¹⁰⁴ = \(\overline{...6}\)

D = 9⁹⁶ = (9²)⁴⁸ = \(\overline{...1}\) ⁴⁸ = \(\overline{...1}\)

E = 2003²⁰⁰⁷ = (2003⁴)⁵⁰¹.2003³ = \(\overline{...1}\) .\(\overline{..7}\) = \(\overline{..7}\)

G = 2022²⁰²² = (2022⁴)⁵⁰⁵.2022² = \(\overline{...6}\).\(\overline{...4}\)  = \(\overline{...4}\)

A = 72008  = (74)1004 = (...1‾...1)1004 = ...1‾...1

B = 23456 = (24)864 = ...6‾...6864 = ...6‾...6

C = 204208 = (2042)104 = ...6‾...6104 = ...6‾...6

D = 996 = (92)48 = ...1‾...1 48 = ...1‾...1

E = 20032007 = (20034)501.20033 = ...1‾...1 ...7‾..7 = ..7‾..7

G = 20222022 = (20224)505.20222 = ...6‾...6....4‾...4  = ...4‾...4

=>\(x^2\)+ \(7x\)=3\(x^2\)+\(7x\)-5

=>\(-2x^2\)+5=0

=>2\(x^2\)-5=0

=>2\(x^2\)=5

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

=>\(x\)=-\(\sqrt{\dfrac{5}{2}}\)

=>\(x\)=+\(\sqrt{\dfrac{5}{2}}\)

`@` `\text {Ans}`

`\downarrow`

\(5^{x-2}-3^2=2^4-\left(6^8\div6^6-6^2\right)\)

`\Rightarrow`\(5^x\div5^2-9=16-\left(6^2-6^2\right)\)

`\Rightarrow`\(5^x\div5^2-9=16\)

`\Rightarrow`\(5^x\div5^2=25\)

`\Rightarrow`\(5^x=5^2\cdot5^2\)

`\Rightarrow`\(5^x=5^4\Rightarrow x=4\)

Vậy, `x = 4.`

5ˣ⁻² - 3² = 2⁴ - (6⁸ : 6⁶ - 6²)

5ˣ⁻² - 9 = 16 - (36 - 36)

5ˣ⁻² - 9 = 16

5ˣ⁻² = 16 + 9

5ˣ⁻² = 25

5ˣ⁻² = 5²

x - 2 = 2

x = 2 + 2

x = 4

3ˣ + 4² = 16

3ˣ + 16 = 16

3ˣ = 16 - 16

3ˣ = 0 (vô lý)

Vậy không tìm được x thỏa mabx yêu cầu

`3^x+4^2=16`

`=>3^x+16=16`

`=>3^x=16-16`

`=>3^x=0`

`=>x=1`

\(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\\ \Leftrightarrow5x^2-15x=5x^2-11x+2-5\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)

\(a)\) Ta có : \(M:N:P:Q=1:2:3:4\)

\(\Rightarrow\dfrac{M}{1}=\dfrac{N}{2}=\dfrac{P}{3}=\dfrac{Q}{4}\left(1\right)\) 

Áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\left(1\right)=\dfrac{M+N+P+Q}{1+2+3+4}=\dfrac{360}{10}=36\)

\(\Rightarrow\left\{{}\begin{matrix}M=36.1=36\\N=36.2=72\\P=36.3=108\\Q=36.4=144\end{matrix}\right.\)

\(b)\) Xét từ giác MNPQ có : \(gócM+gócQ=36+144=180độ\)

Mà : 2 góc ở vị trí trong cùng phía .

\(\Rightarrow MN//PQ\left(đpcm\right)\)

a) \(\dfrac{M}{1}=\dfrac{N}{2}=\dfrac{P}{3}=\dfrac{Q}{4}=\dfrac{M+N+P+Q}{1+2+3+4}=\dfrac{360}{10}=36\)

\(\Rightarrow\left\{{}\begin{matrix}M=36.1=36^o\\N=36.2=72^o\\P=36.3=108^o\\Q=36.4=144^o\end{matrix}\right.\)

  \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)

Vì \(\left(\dfrac{-1}{3}\right)^{-1}=\left(\dfrac{1}{-3}\right)^{-1}=\left(-3^{-1}\right)^{-1}=-3^{-1\times\left(-1\right)}=-3^1=-3\) 

=> \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)