Dư 2 hay dư 2x????

a) Áp dụng BĐT Bunhia ta có:

\(\left(3+1\right)\left(3x^2+y^2\right)\ge\left(3x+y\right)^2\)

<=> \(3x^2+y^2\ge3^2:4=\dfrac{9}{4}\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}3x+y=3\\x=y\end{matrix}\right.\) <=> \(x=y=\dfrac{3}{4}\)

b) Ta có: \(3x+y=3\) => \(y=3-3x\) (1)

Thay (1) vào N ta được:

N = \(2.\left(3-3x\right)x\) = \(6x-6x^2\) = \(-6\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{2}\)

= \(-6\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\) \(\le\) \(\dfrac{3}{2}\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=3-3x\\x=\dfrac{1}{2}\end{matrix}\right.\) <=> \(x=\dfrac{1}{2};y=\dfrac{3}{2}\)

1) a=3k+2=> a+1 chia het cho 3

a=7q+6 => a+1 chia het cho 7

nên a+1 chia hết cho 3.7=21 nên a chia cho 21 dư -1 hay 20

2)a=2p; b=2q với (p;q)=1 mà a+b=10 nên p+q=5 

p=1;q=4 => a=2;b=8

p=2;q=3=> a=4;b=6

a;b có vai trò như nhau.

Nỗi hứng lm cho vui!

Bài 1:

a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)

Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12

=> Dấu = xảy ra <=> \(x=2\)

b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)

= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)

= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)

= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)

Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)

Cách 1: Theo casio ta có:

+ \(\sqrt{3}+\sqrt{7}\approx4,378\)

+ \(\sqrt{19}\approx4,36\)

=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)

Cách 2: Ta có: \(\left(\sqrt{3}+\sqrt{7}\right)^2=3+7+2.\sqrt{21}=10+\sqrt{84}\)

\(\left(\sqrt{19}\right)^2=19=10+\sqrt{81}\)

Vì \(10+\sqrt{84}>10+\sqrt{81}\)

=> \(\left(\sqrt{3}+\sqrt{7}\right)^2>\left(\sqrt{19}\right)^2\)

=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)

Nhân ra có bớt đc em nào đâu thì lm sao mà rút gọn hả trời????