\(a,2^n+2^{n+4}=272\\ \Rightarrow2^n+2^n.2^4=272\\ \Rightarrow2^n+2^n.16=272\\ \Rightarrow2^n.17=272\\ \Rightarrow2^n=16\\ \Rightarrow2^n=2^4\\ \Rightarrow n=4\)

\(b,5^{n+2}-5^n=600\\ \Rightarrow5^n.5^2-5^n=600\\ \Rightarrow5^n\left(25-1\right)=600\\ \Rightarrow5^n.24=600\\ \Rightarrow5^n=25\\ \Rightarrow5^n=5^2\\ \Rightarrow n=2\)

\(a)2^n+2^{n+4}=272\)

\(2^n+2^n.2^4=272\)

\(2^n\left(1+2^4\right)=272\)

\(2^n.17=272\)

\(2^n=16\)

\(2^n=2^4\)

\(\Rightarrow n=4\)

\(b)\)\(5^{n+2}-5^n=600\)

\(5^n.5^2-5^n=600\)

\(5^n\left(5^2-1\right)=600\)

\(5^n.24=600\)

\(5^n=25\)

\(5^n=5^2\)

\(\Rightarrow n=2\)

Chúc bạn học tốt ❤️❤️

a: 3x=2y

=>\(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=10

nen Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\)

=>\(x=2\cdot2=4;y=2\cdot3=6\)

b: \(\dfrac{x-2}{y+3}=\dfrac{8}{12}=\dfrac{2}{3}\)

=>3x-6=2y+6

=>3x-2y=12

y-x=-4

=>x=y-(-4)=y+4

3x-2y=12

=>3(y+4)-2y=12

=>3y+12-2y=12

=>y=0

x=y+4=0+4=4

c: \(\dfrac{x}{2}=\dfrac{y}{5}\)

mà x+2y=12

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+2y}{2+5\cdot2}=\dfrac{12}{12}=1\)

=>x=2;y=5

\(a,27\cdot3^{n-1}=81\\ \Rightarrow3^{n-1}=81:27\\ \Rightarrow3^{n-1}=3\\ \Rightarrow n-1=1\\ \Rightarrow n=2\)

\(b,5\cdot5^{6-n}=625\\ \Rightarrow5^{6-n}=625:5\\ \Rightarrow5^{6-n}=125\\ \Rightarrow5^{6-n}=5^3\\ \Rightarrow6-n=3\\ \Rightarrow n=3\)

\(a,27\cdot3^{n-1}=81\\ =>3^3\cdot3^{n-1}=81\\ =>3^{n-1+3}=3^4\\ =>3^{n+2}=3^4\\ =>n+2=4\\ =>n=4-2\\ =>n=2\\ b,5\cdot5^{6-n}=625\\ =>5^{1+6-n}=5^4\\ =>5^{7-n}=5^4\\ =>7-n=4\\ =>n=7-4\\=>n=3\)

\(\dfrac{x}{3}=\dfrac{2}{5}\\ =>x=3\cdot\dfrac{2}{5}\\ =>x=\dfrac{3\cdot2}{5}\\ =>x=\dfrac{6}{5}\)

Vậy: ... 

`x/3 = 2/5`

`=>5x = 2 . 3`

`=>5x=6`

`=>x=6/5`

Vậy: `x=6/5`

\(a,2^{n-1}=16\\ =>2^{n-1}=2^4\\ =>n-1=4\\ =>n=4+1\\ =>n=5\\ b,3^{21}:3^7:3\\ =3^{21-7-1}\\ =3^{14-1}\\ =3^{13}\)

\(a,9^{11}:9^2:9^3\\ =9^{11-2-3}\\ =9^6\\ =\left(3^2\right)^6\\ =3^{2\cdot6}\\ =3^{12}\\ b,3^{21}:3^7:3\\ =3^{21-7-1}\\ =3^{13}\)

a) (Sửa đề) 
\(9^{11}:9^2:9:3\\ =\left(3^2\right)^{11}:\left(3^2\right)^2:3^2:3\\ =3^{22}:3^4:3^2:3\\ =3^{22}:\left(3^4.3^2.3\right)\\ =3^{22}:3^7\\ =3^{15}\)

b) 
\(3^{21}:3^7:3\\ =3^{21}:\left(3^7.3\right)\\ =3^{21}:3^8\\ =3^{13}\)

\(a,6^8\cdot6^8\cdot6^5\\ =6^{8+8+5}\\ =6^{21}\\ b,25^2\cdot5^3\cdot5\\ =\left(5^2\right)^2\cdot5^3\cdot5\\ =5^{2\cdot2}\cdot5^3\cdot5\\ =5^4\cdot5^3\cdot5\\ =5^{4+3+1}\\ =5^8\)

a) Ta có:
\(x-y⋮7\)
Vì \(21x⋮7\) nên:
\(x-y+21x⋮\\ \Rightarrow22x-y⋮7\)
Vậy...

b) Ta có:
\(x-y⋮7\)
Vì \(7x⋮7\) và \(21y⋮7\) nên:
\(x-y+7x+21y⋮\\ \Rightarrow8x+20y⋮7\)
Vậy...

c) Ta có:
\(x-y⋮7\\ \Rightarrow11.\left(x-y\right)⋮7\\ \Rightarrow11x-11y⋮7\)
Vì \(21y⋮7\) nên:
\(11x-11y+21y⋮\\ \Rightarrow11x+10y⋮7\)
Vậy...

Ta có `x - y ⋮ 7`

`=>x-y=7k(k∈N)` 

`=>x=7k+y`

a) `22x-y`

`=22(7k+y)-y`

`=7k*22+22y-y`

`=7k*22+21y`

`=7*(22k+3y)⋮7` 

b) `8x+20y`

`=8(7k+y)+20y`

`=56k+8y+20y`

`=56k+28y`

`=7*(8k+4y)⋮7`

c) `11x+10y`

`=11(7k+y)+10y`

`=77k+11y+10y`

`=77k+21y`

`=7*(11k+3y)⋮7` 

Gọi số cần tìm là x

Theo đề, ta có: \(\dfrac{-9}{5}< \dfrac{-7}{x}< \dfrac{-9}{6}\)

=>\(\dfrac{9}{5}>\dfrac{7}{x}>\dfrac{9}{6}\)

=>\(\dfrac{63}{35}>\dfrac{63}{9x}>\dfrac{63}{42}\)

=>35<9x<42

=>9x=36

=>x=4

 Gọi mẫu số của phân số cần tìm là \(x\) ( 0 ≠ \(x\) \(\in\) Z)

       \(\dfrac{-9}{5}\) < \(\dfrac{-7}{x}\) < \(\dfrac{-9}{6}\)

    ⇒ \(\dfrac{-9.\left(-1\right)}{5.}\) > \(\dfrac{-7.\left(-1\right)}{x}\) > \(\dfrac{-9.\left(-1\right)}{6}\) (nhân cả hai vế của bất đẳng thức với một số âm thì bất đẳng thức đổi chiều)

 ⇒\(\dfrac{9}{5}\) > \(\dfrac{7}{x}\) > \(\dfrac{9}{6}\)

⇒ \(\dfrac{9.7}{5.7}\) > \(\dfrac{7.9}{x.9}\) > \(\dfrac{9.7}{6.7}\)

⇒ \(\dfrac{63}{35}\) > \(\dfrac{63}{9x}\) > \(\dfrac{63}{42}\)

⇒ 35 < 9\(x\) < 42

      \(\dfrac{35}{9}\) < \(x\) < \(\dfrac{42}{9}\)

      3\(\dfrac{8}{9}\) < \(x\) < 4\(\dfrac{6}{9}\)

Vậy \(x=4\) 

1: \(3^{x+2}+3^x=10\)

=>\(9\cdot3^x+3^x=10\)

=>\(10\cdot3^x=10\)

=>\(3^x=1\)

=>x=0

2: \(2^{x+1}-2^x=32\)

=>\(2^x\cdot2-2^x=32\)

=>\(2^x=32=2^5\)

=>x=5

3: \(4^{x+2}-4^x=60\)

=>\(4^x\cdot16-4^x=60\)

=>\(15\cdot4^x=15\cdot4\)

=>\(4^x=4\)

=>x=1

4: \(2^{x+2}-2^x=96\)

=>\(4\cdot2^x-2^x=96\)

=>\(3\cdot2^x=3\cdot32\)

=>\(2^x=32\)

=>x=5

5: \(2^{x+3}+2^x=144\)

=>\(2^x\cdot8+2^x=144\)

=>\(9\cdot2^x=9\cdot16\)

=>\(2^x=16\)

=>x=4

6: \(3^{x+3}-3^x=234\)

=>\(3^x\cdot27-3^x=234\)

=>\(26\cdot3^x=234\)

=>\(3^x=9=3^2\)

=>x=2

7: 

\(5^x+5^{x+1}=750\)

=>\(5^x+5\cdot5^x=750\)

=>\(6\cdot5^x=750\)

=>\(5^x=125=5^3\)

=>x=3

8: \(2^x+2^{x+2}=320\)

=>\(2^x+2^x\cdot4=320\)

=>\(5\cdot2^x=320\)

=>\(2^x=64=2^6\)

=>x=6

9: \(5^x+5^{x+2}=650\)

=>\(5^x+5^x\cdot25=650\)

=>\(26\cdot5^x=650\)

=>\(5^x=\dfrac{650}{26}=25=5^2\)

=>x=2