Ta có:

\(\left(-16\right)^{11}=-16^{11}=-\left(2^4\right)^{11}=-2^{4\cdot11}=-2^{44}\)

\(\left(-32\right)^9=-32^9=-\left(2^5\right)^9=-2^{5\cdot9}=-2^{45}\)

Mà: \(44< 45\)

\(\Rightarrow2^{44}< 2^{45}\)

\(\Rightarrow-2^{44}>-2^{45}\)

${\left(-16\right)}^{11}=-16^{11}=-{\left(2^4\right)}^{11}=-2^{4\cdot 11}=-2^{44}$

${\left(-32\right)}^9=-32^9=-{\left(2^5\right)}^9=-2^{5\cdot 9}=-2^{45}$

Mà                 44<4544<45

$\Rightarrow 2^{44}<2^{45}$

$\Rightarrow -2^{44}>-2^{45}$

\(3^{1001}\cdot\left(5+2x\right)=3^{999}\)

\(\Rightarrow5+2x=\dfrac{3^{999}}{3^{1001}}\)

\(\Rightarrow5+2x=\dfrac{1}{3^2}\)

\(\Rightarrow5+2x=\dfrac{1}{9}\)

\(\Rightarrow2x=\dfrac{1}{9}-5\)

\(\Rightarrow2x=-\dfrac{44}{9}\)

\(\Rightarrow x=-\dfrac{44}{9}:2\)

\(\Rightarrow x=-\dfrac{22}{9}\)

\(3^{1001}.\left(5+2x\right)=3^{999}\)

\(\Rightarrow5+2x=\dfrac{3^{999}}{3^{1001}}\)

\(\Rightarrow5+2x=\dfrac{1}{3^2}\)

\(\Rightarrow5+2x=\dfrac{1}{9}\)

\(2x=\dfrac{1}{9}-5\)

\(x=-\dfrac{44}{9}:2\)

\(x=-\dfrac{22}{9}\)

\(\dfrac{5}{4}-\dfrac{1}{4}\left(x-\dfrac{2}{3}\right)=\dfrac{5}{2}\\ =>\dfrac{1}{4}\left(x-\dfrac{2}{3}\right)=\dfrac{5}{4}-\dfrac{5}{2}=-\dfrac{5}{4}\\ =>x-\dfrac{2}{3}=-\dfrac{5}{4}:\dfrac{1}{4}=-5\\ =>x=-5+\dfrac{2}{3}=-\dfrac{13}{3}\)

a)

\(3^{21}-3^{18}\\ =3^{17}.\left(3^4-3\right)\\ =3^{17}.\left(81-3\right)\\ =3^{17}.78\)

Vì \(3^{17}.78⋮78\) nên \(3^{21}-3^{18}⋮78\) (đpcm)

Vậy...

b)
\(81^7-27^9-9^{13}\\ =\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\\ =3^{28}-3^{27}-3^{26}\\ =3^{24}.\left(3^4-3^3-3^2\right)\\ =3^{24}.\left(81-27-9\right)\\ =3^{24}.45\)

Vì \(3^{24}.45⋮45\) nên \(81^7-27^9-9^{13}⋮45\) (đpcm)

Vậy...

\(2^{n+1}=2^3\)

\(\Rightarrow n+1=3\)

\(\Rightarrow n=3-1\)

\(\Rightarrow n=2\)

Vậy: n=2 

\(\dfrac{7}{5}.x+\dfrac{4}{5}=\dfrac{-7}{10}\)

\(=>\dfrac{7}{5}.x=\dfrac{-7}{10}-\dfrac{4}{5}=\dfrac{-7}{10}-\dfrac{8}{10}\)

\(=>\dfrac{7}{5}.x=-\dfrac{15}{10}\)

\(=>x=\dfrac{-15}{10}:\dfrac{7}{5}=\dfrac{-15}{10}.\dfrac{5}{7}\)

\(=>x=\dfrac{-15}{14}\)

\(\frac75\cdot x+\frac45=\frac{-7}{10}\\\Rightarrow \frac75\cdot x=\frac{-7}{10}-\frac45\\\Rightarrow \frac75\cdot x=-\frac32\\\Rightarrow x=-\frac32:\frac75\\\Rightarrow x=-\frac{15}{14}\)

`#3107.101107`

\(\dfrac{1}{4}\times x+\dfrac{4}{5}=-\dfrac{7}{10}\)

\(\Rightarrow\dfrac{1}{4}\times x=-\dfrac{7}{10}-\dfrac{4}{5}\)

\(\Rightarrow\dfrac{1}{4}\times x=-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{3}{2}\div\dfrac{1}{4}\)

\(\Rightarrow x=-6\)

Vậy, `x = -6.`

1/4 . x + 4/5 = -7/10

1/4 . x = -7/10 - 4/5

1/4 . x = -3/2

x = -3/2 : 1/4

x = -6

a) Do BD là tia phân giác của ABC (gt)

⇒ ∠ABD = ∠CBD = 90⁰ : 2 = 45⁰

Do MN // AB (gt)

AB ⊥ BC (gt)

⇒ MN ⊥ BC

⇒ ∠INB = 90⁰

⇒ ∆INB vuông tại N

⇒ ∠BIN = 90⁰ - ∠IBN

= 90⁰ - ∠CBD

= 90⁰ - 45⁰

= 45⁰

⇒ ∠MID = ∠BIN = 45⁰ (đối đỉnh)

b) MN ⊥ BC (đã chứng minh ở câu a)

Tính MID MÀ