1. This operation has never been_____in this country

A. preserved

B.performed

C.worshipped

D.commemorated

2. I've got no sense of rhythm, so I'm a terrible____

a.dancer  B.parade  C. crowd.   D.procession

3. He didn't want to go to the dentist,_____he went anyway

A.and  B.for   C.yet   D. and

4. We always_____our wedding anniversary by going out to dinner

A. commemorate   B. remember    C.worship   D. celebrate

1. This operation has never been_____in this country

A. preserved

B.performed

C.worshipped

D.commemorated

2. I've got no sense of rhythm, so I'm a terrible____

a.dancer  B.parade  C. crowd.   D.procession

3. He didn't want to go to the dentist,_____he went anyway

A.and  B.for   C.yet   D. and

4. We always_____our wedding anniversary by going out to dinner

A. commemorate   B. remember    C.worship   D. celebrate

Bài làm

Ta có : 25n⁴ + 50n³ - n² - 2n

= 24n⁴ + n⁴ + 48n³ + 2n³ - n² - 2n

= ( 24n⁴ + 48n³ ) + ( n⁴ + 2n³ - n² - 2n )

= 24n³( n + 2 ) + n( n³ + 2n² - n - 2 )

= 24n³( n + 2 ) + n[ n²( n + 2 ) - 1( n + 2 ) ]

= 24n³( n + 2 ) + n( n + 2 )( n² - 1 )

= 24n³( n + 2 ) + ( n - 1 )n( n + 1 )( n + 2 )

Dễ dàng chứng minh ( n - 1 )n( n + 1 )( n + 2 ) chia hết cho 24

Vì \(\hept{\begin{cases}\left[24n^3\left(n+2\right)\right]⋮24\\\left[\left(n-1\right)n\left(n+1\right)\left(n+2\right)\right]⋮24\end{cases}}\Rightarrow\left[24n^3\left(n+2\right)+\left(n-1\right)n\left(n+1\right)\left(n+2\right)\right]⋮24\)

hay ( 25n⁴ + 50n³ - n² - 2n ) chia hết cho 24 ( đpcm )

\(\frac{1}{x}-\frac{1}{2y}=\frac{1}{2x+y}\)

=> \(\frac{2y-x}{2xy}=\frac{1}{2x+y}\)

=> (2y - x)(2x + y) = 2xy

=> 4xy + 2y² - 2x² - xy = 2xy

=> 2(y²- x²) = -xy

=> [2(y² - x²)]² = (-xy)²

=> 4(y² - x²)² = (xy)²

=> 4(y⁴ - 2(xy)² + x⁴) = (xy)²

=> 4y⁴ - 8(xy)² + 4x⁴ = (xy)²

=> 4(y⁴ + x⁴) = 9(xy)²

=> y⁴ + x⁴ = \(\frac{9}{4}\left(xy\right)^2\)

Khi đó  \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=\frac{x^4+y^4}{\left(xy\right)^2}=\frac{\frac{9}{4}\left(xy\right)^2}{\left(xy\right)^2}=\frac{9}{4}\)

Ta có :

\(x+\frac{1}{x}=3\)

\(\Rightarrow\left(x+\frac{1}{x}\right)^4=3^4\)

\(\Rightarrow x^4+\frac{1}{x^4}+4x^2+6+\frac{4}{x^2}=81\)(1)

Lại có :

\(4\left(x+\frac{1}{x}\right)^2=4\left(3\right)^2\)

\(\Rightarrow4x^2+8+\frac{4}{x^2}=36\)

\(\Rightarrow4x^2+6+\frac{4}{x^2}=34\)(2)

Thay (2) vào (1) ta có : \(x^4+\frac{1}{x^4}+34=81\)

\(\Rightarrow R=x^4+\frac{1}{x^4}=81-34=47\)

Xem lại đề bạn nhé