\(=3^2.3^n-2^2.2^n+3^n-2^n=10.3^n-5.2^n=\)

\(=10.3^n-10.2^{n-1}=10.\left(3^n-2^n\right)⋮10\)

\(\left(3b^2\right)^2-b^3.\left(1-5b\right)\)

\(=9b^4-b^3+5b^4\)

\(=14b^4-b^3\)

4x\(^{1+n-2}\)y\(^{1+n+1}\)4xy\(^{1+n}\)+4xy

\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)

\(21⋮7\Rightarrow\left(5^3.21\right)⋮7\Rightarrow\left(5^5-5^4+5^3\right)⋮7\)

`@` `\text {Ans}`

`\downarrow`

\(x^n\cdot y^{n+2}\cdot\left(xy+x^2y+1\right)\)

`=`\(x^n\cdot y^n\cdot y^2\left(xy+x^2y+1\right)\)

`=`\(\left(xy\right)^n\cdot y^2\cdot xy+y^2\cdot x^2y+y^2\)

`=`\(\left(xy\right)^n\cdot xy^3+x^2y^3+y^2\)

x\(^{n+1}\)y\(^{n+1}\)+2xy+x\(^{ }\)\(^{n+2}\)y\(\)\(^{n+1}\)+2x\(^2\)y+x\(^n\)y\(^n\)+2

Không: 2 góc đối đỉnh gồm 2 góc có chung một đỉnh, mỗi cạnh của góc này là tia đối của góc kia. 2 góc bẹt nó chỉ tạo nên một góc thôi nhé!

Có vì hai góc bẹt cắt nhau sẽ thành hai gó đối đỉnh

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{\left(x-17\right)\times725}{33\times725}+\dfrac{\left(x-21\right)\times825}{29\times825}+\dfrac{x\times957}{25\times957}-\dfrac{4\times23925}{23925}=0\)

\(725x-12325+825x-17325+957x-95700=0\)

\(2507x-125350=0\)

\(2507x=125350\)

\(x=50\)

Nếu mà theo cách x - 50 = 0 thì bạn theo cách này nha:

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{x-17}{33}-1+\dfrac{x-21}{29}-1+\dfrac{x}{25}-2=0\)

\(\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)

\(\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{5}\right)=0\)

Vì  \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}>0\)

=> \(x-50=0\)
=> \(x=50\)