Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
! có gì mình tick cho nha!
\(\Leftrightarrow n^2+4n+3n+12-10⋮n+4\)
\(\Leftrightarrow n+4\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(n\in\left\{1;6\right\}\)
1,=0 . [2017/2018+2018/2019]
=>0
2,TH1 x-3=0=>x=3
TH2 y-4=0=>y=4
3, -2/4 = -x/10 = 16/y
=>-1/2 = -x/10 = 16/y
=>-1/2 = -x/10 => -5/10 = -x/10 => x=5
-1/2 = 16/y => 16/-32 = 16/y => y = -32
a) Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in Z\right)\)
\(\Rightarrow B=\dfrac{5^{12}+2}{5^{13}+2}< 1\)
\(B< \dfrac{5^{12}+2+48}{5^{13}+2+48}\Rightarrow B< \dfrac{5^{12}+50}{5^{13}+50}\Rightarrow B< \dfrac{5^2\left(5^{10}+2\right)}{5^2\left(5^{11}+2\right)}\Rightarrow B< \dfrac{5^{10}+2}{5^{11}+2}=A\)\(B< A\)
bạn ơi thế còn phần b thì sao? Mong bạn có câu trả lời sớm tớ cảm ơn bạn nhiều lắm
a/ Ta có :
\(8^5=\left(2^3\right)^5=2^{15}=2.2^{14}\)
\(3.4^7=3.\left(2^2\right)^7=3.2^{14}\)
Vì \(2.2^{14}< 3.2^{14}\Leftrightarrow8^5< 3.4^7\)
b/ Ta có :
\(3^{21}=3^{20}.3=\left(3^2\right)^{10}.3=9^{10}.3\)
\(2^{31}=2^{30}.2=\left(2^3\right)^{10}.2=8^{10}.2\)
Vì \(9^{10}.3>8^{10}.2\Leftrightarrow3^{21}>2^{31}\)
a) \(\dfrac{n}{3n+1}=\dfrac{2.n}{2\left(3n+1\right)}=\dfrac{2n}{6n+2}\)
Vì \(\dfrac{2n}{6n+2}< \dfrac{2n}{6n+1}\Leftrightarrow\dfrac{n}{3n+1}< \dfrac{2n}{6n+1}\)
b) Áp dụng công thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a;b;m\in N\cdot\right)\)
Ta có :
\(B=\dfrac{10^8+1}{10^9+1}< 1\)
\(\Leftrightarrow B=\dfrac{10^8+1}{10^9+1}< \dfrac{10^8+1+9}{10^9+1+9}=\dfrac{10^8+10}{10^9+10}=\dfrac{10\left(10^7+1\right)}{10\left(10^8+1\right)}=\dfrac{10^7+1}{10^8+1}=A\)
\(\Leftrightarrow B< A\)
Ta có:
\(\dfrac{n}{3n+1}=\dfrac{2n}{2\left(3n+1\right)}=\dfrac{2n}{6n+2}\)
\(\dfrac{2n}{6n+2}< \dfrac{2n}{6n+1}\Rightarrow\dfrac{n}{3n+1}< \dfrac{2n}{6n+1}\)
Ta có:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^8+1}{10^9+1}< 1\)
\(\Rightarrow B< \dfrac{10^8+1+9}{10^9+1+9}\Rightarrow B< \dfrac{10^8+10}{10^9+10}\Rightarrow B< \dfrac{10\left(10^7+1\right)}{10\left(10^8+1\right)}\Rightarrow B< \dfrac{10^7+1}{10^8+1}=A\)\(\Rightarrow B< A\)
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
không biết đúng hay sai mà cũng cảm ơn bạn đã trả lời giúp mình !
