a)A=\(1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)

Vậy A < B

b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=A\)

Vậy B < A

\(x^2-2x=24\)

\(x^2-2x-24=0\)

\(x^2+4x-6x-24=0\)

\(x\left(x+4\right)-6\left(x+4\right)=0\)

\(\left(x+4\right)\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)

A = \(\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)\)

A = \(x^2-6x+9-4x^2+1=-3x^2-6x+10\)

B = \(\left(2x-3\right)^2-\left(x-1\right)\left(2x+1\right)\)

B = \(4x^2-12x+9-2x^2-x+2x+1\)

B = \(2x^2-11x+10\)

C = \(4x\left(x-3\right)^2-\left(4-2x\right)^2\)

C = \(4x\left(x^2-6x+9\right)-16+16x-4x^2\)

C = \(4x^3-24x^2+36x-16+16x-4x^2\)

C = \(4x^3-28x^2+52x-16\)

D = \(3x\left(x-1\right)\left(x-2\right)-x\left(2x-1\right)^2\)

D = \(\left(3x^2-3x\right)\left(x-2\right)-x\left(2x-1\right)^2\)

D = \(3x^3-6x^2-3x^2+6x-x\left(4x^2-4x+1\right)\)

D = \(3x^3-9x^2+6x-4x^3+4x^2-x\)

D = \(-x^3-5x^2+5x\)

\(\left(2x+1\right)^2-\left(1-2x\right)^2=16\)

\(\left(2x+1-1+2x\right)\left(2x+1+1-2x\right)=16\)

\(4x.2=16\)

\(8x=16\Rightarrow x=2\)

M = \(\left(x-1\right)^3-x\left(x+2\right)^2+7x\left(x+\dfrac{1}{7}\right)\)

M=\(x^3-3x^2+3x-1-x.\left(x^2+4x+4\right)+7x^2+x\)

M = \(x^3-3x^2+3x-1-x^3-4x^2-4x+7x^2+x\)

M = \(-1\)

Vậy...

1) Ta có:

\(2x-x^2-3=-\left(x^2-2x+3\right)\)

= \(-\left(x^2-2x+1+2\right)\)

= \(-\left[\left(x+1\right)^2+2\right]\)

= \(-\left(x+1\right)^2-2< 0\) với mọi x ( đpcm )

Ta có : \(n^4-4n^3-4n^2+16n\)

= \(n^3\left(n-4\right)-4n\left(n-4\right)\)

= \(\left(n-4\right)\left(n^3-4n\right)=\left(n-4\right)n\left(n^2-4\right)\)

= \(\left(n-4\right).n.\left(n-2\right)\left(n+2\right)\)

= \(\left(n-4\right).\left(n-2\right).n.\left(n+2\right)\)

Dấu hiệu chia hết cho 384: Tích của 4 số chẵn liên tiếp thì chia hết cho 384.

Ta thấy kết quả \(\left(n-4\right).\left(n-2\right).n.\left(n+2\right)\) vốn đã là tích của 4 số chẵn liên tiếp, do đó tích trên chia hết cho 384 với mọi n chẵn và n > 4

B1:

a) \(x^2\left(x+2\right)-3x\left(x^2-1\right)\)

= \(x^3+2x^2-3x^3+3x=-2x^3+2x^2+3x\)

b) \(x^3\left(x-4\right)+\left(x^2+3\right)\left(-x\right)-x^4\)

= \(x^4-4x^3-x^3-3x-x^4=-5x^3-3x\)

B2:

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2-26=0\)

\(-13x-26=0\)

\(-13\left(x+2\right)=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

b) \(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5=4\)

\(x^3+x^2+x-x^2-x^3-x^2-x+5-4=0\)

\(-x^2+1=0\Rightarrow-x^2=-1\Rightarrow x=\pm1\)

-Dùng từ gợi ý viết thành câu hoàn chỉnh:

1/Mrs.Xuan/say/she/be/happy/tosee you.

=> Mr.s Xuan said she was happy to see you.

2/Christopher Columbus/be/the fist explorer/who/discover/America.

=> Christopher Columbus is the first explorer who discovered America

-Supply the corret form or tense of the verb:

1/...Is......Nam(go)....going....to West Lake Park on this birthday?

2/Mai still in her room. She (iron)..is ironing......her clothes for next week.