a.5 mũ n =5 mũ 78 : 5 mũ 14=5 mũ 78-14=5 mũ 64

a) \(5^n\cdot5^{14}=5^{78}\)

\(\Rightarrow5^{n+14}=5^{78}\)

\(5^{n+14}=5^{64+14}\)

⇒ n + 14 = 64 + 14

⇒ n = 64.

Giải:
Ta có: (n^2+4⋮n+1)

(Rightarrow n^2+n+4-n⋮n+1)

(Rightarrow nleft(n+1 ight)+4-n⋮n+1)

(Rightarrow4-n⋮n+1)

(Rightarrow5-left(n+1 ight)⋮n+1)

(Rightarrow5⋮n+1)

(Rightarrow n+1inleft{1;-1;5;-5 ight})

(Rightarrow ninleft{0;-2;4;-6 ight})

Vậy (ninleft{0;-2;4;-6 ight})

a) Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in Z\right)\)

\(\Rightarrow B=\dfrac{5^{12}+2}{5^{13}+2}< 1\)

\(B< \dfrac{5^{12}+2+48}{5^{13}+2+48}\Rightarrow B< \dfrac{5^{12}+50}{5^{13}+50}\Rightarrow B< \dfrac{5^2\left(5^{10}+2\right)}{5^2\left(5^{11}+2\right)}\Rightarrow B< \dfrac{5^{10}+2}{5^{11}+2}=A\)\(B< A\)

a) đúng

b) sai

c) sai

d) đúng

khẳng định a) đúng

khẳng định b) sai

khẳng định c) sai

khẳng định d) đúng

Câu hỏi của nguyen thu thi - Toán lớp 6 | Học trực tuyến

cau hoc truong nnao ma de the ko biet lam

1.ta có : \(\frac{x}{14}\)=\(\frac{6}{y}\)=\(\frac{-15}{35}\)

\(\Rightarrow\)\(\frac{x}{14}\)=\(\frac{-15}{35}\)

\(\Rightarrow\) x . 35=14 . (-15)

\(\Rightarrow\) x . 35= -210

\(\Rightarrow\) x = -210 : 35

\(\Rightarrow\) x = -6 (thỏa mãn)

\(\Rightarrow\) \(\frac{6}{y}\)=\(\frac{-6}{14}\)

\(\Rightarrow\) y . (-6)= 6 .14

\(\Rightarrow\) y . (-6)= 84

\(\Rightarrow\) y = 84 : (-6)

\(\Rightarrow\) y = -14(thỏa mãn)

Vậy x = -6 ; y = -14

1)

a)\(2^n.16=128\)

\(\Rightarrow2^n=128:16=8\)

\(\Rightarrow2^n=2^3\)

\(\Rightarrow n=3\)

b)\(3^n:9=27\)

\(\Rightarrow3^n=27.9=243\)

\(\Rightarrow3^n=3^5\)

\(\Rightarrow n=5\)

c)\(\left(2n+1\right)^3=27\)

\(\Rightarrow2n+1=3\)

\(\Rightarrow2n=2\)

\(\Rightarrow n=1\)

d)\(\left(n-2\right)^2=\left(n-2\right)^4\)

TH1 : \(n-2=0\Rightarrow n=2\)

TH2:\(n-2=1\Rightarrow n=3\)

2)

a) \(2^{30}\) và \(3^{20}\)

Ta có: \(2^{30}=\left(2^3\right)^{10}=8^{10}\)

\(3^{20}=\left(3^2\right)^{10}=9^{10}\)

\(\Rightarrow8^{10}< 9^{10}\)

\(\Rightarrow2^{30}< 3^{20}\)

b)Tương tự

Ta có: \(\frac{-11}{3}=\frac{-121}{33}\)

\(\frac{22}{11}=\frac{66}{33}\)

\(-4=\frac{-132}{33}\)

=> \(-4\notin A\)

I'm chiu.... ai ma ra de kho du vay

\(\Leftrightarrow n^2+4n+3n+12-10⋮n+4\)

\(\Leftrightarrow n+4\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(n\in\left\{1;6\right\}\)