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

a) (25u²v - 13uv² + u³) - M = 11u²v - 2u²

<=> 25u²v - 13uv² + u³ - 11u²v + 2u² = M

<=> 14u²v - 13uv² + u³ + 2u² = M

<=> M = 14u²v - 13uv² + u³ + 2u²

b) (7xyz + 15x²yz² - 2xy³) + M = 0

<=> M = -7xyz - 15x²yz² + 2xy³

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

dùm mink câu này vs

a) Đặt A = u² + v² - 2u + 3v + 15

= (u² - 2u + 1) + (v² + 3v + 9/4) + 47/4

= (u - 1)² + (v + 3/2)² + 47/4 \(\ge\frac{47}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}u-1=0\\v+\frac{3}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}u=1\\v=-\frac{3}{2}\end{cases}}\)

Vậy Min A = 47/4 <=> u = 1 ; y = -3/2

lô

xem lại đề bạn ơi. nếu( u+2v+1)+(2u-2v+2)=3u+3 và chưa chắc cái này đã lẻ

đề đúng mà

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

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

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

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

a) \(P=y^2+8y+15\)

\(P=y^2+2.y.4+16-1\)

\(P=\left(y+4\right)^2-1\)

Vì \(\left(y+4\right)^2\ge0\) với mọi y

\(\Rightarrow\left(y+4\right)^2-1\ge-1\) với mọi y

\(\Rightarrow Pmin=-1\Leftrightarrow y=-4\)

b) \(A=u^2+v^2-2u+3v+15\)

\(A=u^2-2u+1+v^2+2.v.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}+14\)

\(A=\left(u-1\right)^2+\left(v+\dfrac{3}{2}\right)^2+\dfrac{47}{4}\)

Vì \(\left(u-1\right)^2\ge0\) với mọi u

\(\left(v+\dfrac{3}{2}\right)^2\ge0\) với mọi v

\(\Rightarrow\left(u-1\right)^2+\left(v+\dfrac{3}{2}\right)^2\ge0\) với mọi u,v

\(\Rightarrow\left(u-1\right)^2+\left(v+\dfrac{3}{2}\right)^2+\dfrac{47}{4}\ge\dfrac{47}{4}\)

\(\Rightarrow Amin=\dfrac{47}{4}\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=-\dfrac{3}{2}\end{matrix}\right.\)

\(2u\left(1+u-v\right)-v\left(1-2y+v\right)\)

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

\(=u+2y^2-v^2\)