a) 4x^2(5x^3 - 2x + 3)

= 20x^5 - 8x^3 + 12x^2

b) 2u(1 + u - v) - v(1 - 2u + v)

= 2u + 2u^2 - v - v^2

bn lấy y ở đâu ra thế?

a, \(x^2+2x\left(y+1\right)+y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1\)

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

b, \(u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)

\(=u^2+v^2+2u+2v+2uv+2u+2v+2+2\)

\(=\left(u^2+2uv+v^2\right)+\left(4u+4v\right)+4\)

\(=\left(u+v\right)^2+4\left(u+v\right)+2^2=\left(u+v+2\right)^2\)

1.

a) \(A=x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

b) \(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2\)\(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+1+1\)\(B=\left(u^2+2u+1\right)+2\left(u+1\right)\left(v+1\right)+\left(v^2+2v+1\right)\)\(B=\left(u+1\right)^2+2\left(u+1\right)\left(v+1\right)+\left(v+1\right)^2\)\(B=\left(u+1+v+1\right)^2=\left(u+v+2\right)^2\)

tik mik nha !!!

a.) \(A=x^2+y^2+1+2xy+2x+2y=\left(x+y+1\right)^2.\)

b.) \(B=u^2+v^2+2u+2v+2\left(u+1\right)\left(v+1\right)+2=u^2+2u+1+2\left(u+1\right)\left(v+1\right)+v^2+2v+1\)

\(B=\left(u+1\right)^2+2\left(u+1\right)\left(v+1\right)+\left(v+1\right)^2=\left(u+1+v+1\right)^2=\left(u+v+2\right)^2\)

Giả sử số tự nhiên a chia cho 7 dư 3. CMR a chia cho 7 dư 2

a) Rút gọn I = s 3   +   t 3  Þ I = 0.

b) Rút gọn N = u 3   – v 3  Þ N = 0.

1.

a. \(x^2-4x\Rightarrow x^2-4x+4=\left(x-2\right)^2\)

b. \(x^2+9\Rightarrow x^2+9+6x=\left(x+3\right)^2\)

c. \(x^2+xy+y^2\Rightarrow x^2+xy+y^2+xy=\left(x+y\right)^2\)

d. \(x^2-x\Rightarrow x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

Là sao

Muốn gõ công thức thì bạn gõ vào chữ M nằm ngang mà hiển thì trong phần tl câu hỏi í

a) 3 u 2 − 8 u + 3 ( u 2 + 1 ) ( u − 1 )                  b) 1 − 4 u 4 ( 4 u + 1 )

Giải:

Ta có: \(U_{n-1}=\dfrac{3U_n-U_{n+1}}{2}\) nên:

\(U_4=340;U_3=216;U_2=154;U_1=123\)

Từ \(U_5=588;U_6=1084;U_{n+1}=3U_n-2U_{n-1}\)

\(\Rightarrow\) \(U_{25}=520093788\)

Vậy \(U_2=154;U_1=123;\) \(U_{25}=520093788\)