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

a. A = | x + 2 | + | x - 3 | - 7

=> A = | x + 2 | + | 3 - x | - 7

Áp dụng BĐT | a | + | b |\(\ge\)| a + b | , ta có :

A = | x + 2 | + | 3 - x | - 7 \(\ge\) | x + 2 + 3 - x | - 7 = | 5 | - 7 = 5 - 7 = - 2

Dấu "=" xảy ra <=> \(3\ge x\ge-2\)

Vậy minA = - 2 <=> \(3\ge x\ge-2\)

c. C = - x² + 6x - 4y² - 4y + 5

=> C = - ( x² - 6x + 9 ) - ( 4y² + 4y + 1 ) + 15

=> C = - ( x - 3 )² - 4 ( y - 1/2 )² + 15\(\le\)15\(\forall\)x

Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x-3\right)^2=0\\4\left(y-\frac{1}{2}\right)^2=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

Vậy maxC = 15 <=>\(\orbr{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

Bài làm

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

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

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

\(=\frac{3}{x-1}-\frac{2}{x-1}=\frac{1}{x-1}\)

Bài làm 

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

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

\(=\frac{1-x}{\left(1-x\right)\left(x-1\right)}=\frac{1}{x-1}\)