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

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

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

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

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

\(2^n.17=272\)

\(2^n=16\)

\(2^n=2^4\)

\(\Rightarrow n=4\)

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

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

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

\(5^n.24=600\)

\(5^n=25\)

\(5^n=5^2\)

\(\Rightarrow n=2\)

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đúng vậy

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` 

\(20-4\sqrt{5}x+x^2\\ =x^2-4\sqrt{5}x+20\\ =x^2-2\cdot x\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2\\ =\left(x-2\sqrt{5}\right)^2\)