Tìm x ∈ N

a) 2x chia hết cho 12 ⇒ 2x ∈ B(12) 

2x chia hết cho 30 ⇒ 2x ∈ B(30) 

Mà x có hai chữ số ⇒ 10 ≤ x ≤ 99 

\(\Rightarrow2x\in BC\left(12;30\right)\)

Mà: \(B\left(12\right)=\left\{0;12;24;36;48;60;72;84;96;108;...\right\}\)

\(B\left(30\right)=\left\{0;30;60;90;120;...\right\}\)

\(\Rightarrow BC\left(12;30\right)=\left\{0;60;...\right\}\)

\(\Rightarrow2x=60\)

\(\Rightarrow x=\dfrac{60}{2}\\ \Rightarrow x=30\)

b) \(9^{x+2}-9^{x+1}+9^x=657\)

\(\Rightarrow9^x\cdot\left(9^2-9+1\right)=957\)

\(\Rightarrow9^x\cdot\left(81-8\right)=657\)

\(\Rightarrow9^x\cdot73=657\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)

bạn có thể giải giùm mk bài tính nhanh đc ko??? Mk đang cần gấp á. Cảm ơn bạn nhiều nha!

\(\frac{252}{y}=\frac{84}{5}\)ta có tử số : 252 : 84 = 3 | tức là Ps\(\frac{84}{5}=\frac{84\times3}{5\times3}=\frac{252}{15}\left(y=15\right)\)

\(y-\frac{2}{255}=\frac{38}{51}\)

\(y=\frac{38}{51}-\frac{2}{255}=\frac{188}{255}\)

              Bài làm

a) 252​​/y=84/5

Vì 252 : 3 = 84

=> y : 3 =5 

      y      = 5 x 3

      y      = 15

Vậy y = 15

b) y - 2/225  = 38/51

     y             = 38/51 + 2/225

     y             = 2884/3825

  Vậy y = 2884/3825

= 4/45

4/45 bạn nha

a ) x/2=y/5 suy ra x/4=y/10

x/4=z/3 suy ra x/4=2z/3

suy ra x/4=y/10=2z/3=x+y-2z/4+10-6=8/8=1

x/4=1 suy ra x=1*4=4

y/10=1 suy ra y=10*1=10

z/3=1 suy ra z=3*1=3

\(y+\frac{23}{45}=\frac{51}{85}\Rightarrow y=\frac{51}{85}-\frac{23}{45}\Rightarrow y=\frac{4}{45}\)

51/85=y+23/45

      y =51/85-23/45

      y =2295/3825-1955/3825

      y =340/3825=4/45

=>x(0,2+0,5+0,25+1)=45

=>x*1,95=45

=>x=300/13

a: \(4^8\cdot2^{20}=\left(2^2\right)^8\cdot2^{20}=2^{16}\cdot2^{20}=2^{16+20}=2^{36}\)

\(9^{12}\cdot27^5\cdot81^4=\left(3^2\right)^{12}\cdot\left(3^3\right)^5\cdot\left(3^4\right)^4\)

\(=3^{24}\cdot3^{15}\cdot3^{16}=3^{24+15+16}=3^{55}\)

\(64^3\cdot4^5\cdot16^2=\left(4^3\right)^3\cdot4^5\cdot\left(4^2\right)^2=4^9\cdot4^5\cdot4^4=4^{9+5+4}=4^{18}\)

b: \(25^{20}\cdot125^4=\left(5^2\right)^{20}\cdot\left(5^3\right)^4=5^{40}\cdot5^{12}=5^{52}\)

\(x^7\cdot x^4\cdot x^3=x^{7+4+3}=x^{14}\)

\(3^6\cdot4^6=\left(3\cdot4\right)^6=12^6\)

c: \(8^4\cdot2^3\cdot16^2=\left(2^3\right)^4\cdot2^3\cdot\left(2^4\right)^2\)

\(=2^{12}\cdot2^3\cdot2^8=2^{23}\)

\(y\cdot y^7=y^{1+7}=y^8\)

\(2^3\cdot2^2\cdot8^3=2^5\cdot\left(2^3\right)^3=2^5\cdot2^9=2^{5+9}=2^{14}\)

=4/45

ai k mk mk k lai cho

y=4/45 pạn ạ