Bài 1:

a: \(\overrightarrow{AB}\cdot\overrightarrow{BC}=-\overrightarrow{BA}\cdot\overrightarrow{BC}\)

\(=-BA\cdot BC\cdot cosABC=-a^2\cdot cos90=0\)

ABCD là hình vuông

=>\(BD^2=AB^2+AD^2=a^2+a^2=2a^2\)

=>\(BD=a\sqrt2\)

\(\overrightarrow{AB}\cdot\overrightarrow{BD}=-\overrightarrow{BA}\cdot\overrightarrow{BD}\)

\(=-BA\cdot BD\cdot cosABD\)

\(=-a\cdot a\sqrt2\cdot cos45=-a^2\sqrt2\cdot\frac{\sqrt2}{2}=-a^2\)

\(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)

\(=\overrightarrow{AC}\cdot\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\cdot\overrightarrow{BD}+\overrightarrow{AC}\cdot\overrightarrow{BC}\)

\(=\overrightarrow{CB}\cdot\overrightarrow{CA}=CB\cdot CA\cdot cosACB=a\cdot a\sqrt2\cdot cos45=a^2\)

đơn giản!!!

a) 18, 42

18 = 2 . 3\(^2\)

42 = 2 . 3 . 7

ƯCLN(18, 42) = 2 . 3\(^2\)= 18

b) 28, 48

28 = 2\(^2\). 7

48 = 2\(^4\). 3

ƯCLN(28, 48) = 2\(^4\)= 16

chúc may mắn!

Số lớn là:

\(42\div3\times11=154\)

Đáp số: \(154\)

Mọi người tk mình đi mình đang bị âm nè!!!!!!Ai tk mình mình tk lại nha !!!

A=(1/30+1/42+1/56+1/72+1/90+1/110+1/132).2

= (1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12).2

=(1/5-1/12).2

=7/60.2

=14/60=7/30

1, Ta có: \(A=3x^2+8x+9=3\left(x^2+\frac{8}{3}x+3\right)=3\left(x^2+\frac{8}{3}x+\frac{16}{9}+\frac{11}{9}\right)\)

\(=3\left(x+\frac{4}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\forall x\)

=> Min A = 11/3 tại x = -4/3

2, Ta có: \(A=-2x^2+6x+3=-2\left(x^2-3x-\frac{3}{2}\right)=-2\left(x^2-3x+\frac{9}{4}-\frac{15}{4}\right)\)

\(=-2\left(x-\frac{3}{2}\right)^2+\frac{15}{2}\le\frac{15}{2}\forall x\)

=> Max A = 15/2 tại x = 3/2

=.= hk tốt!!

Cảm ơn

a) 58

b) 27

em ơi , (chị lớp 6 ahihi) a là 58 b là 27 em nhé 

a, Ta có: \(A=\left|x+2\right|+\left|9-x\right|\ge\left|X+2+9-x\right|=11\)

Dấu "=' xảy ra khi \(\left(x+2\right)\left(9-x\right)\ge0\Leftrightarrow-2\le x\le9\)

Vậy Min_A = 11 khi -2 =< x =< 9

b, Vì \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow B=\frac{3}{4}-\left(x-1\right)^2\le\frac{3}{4}\)

Dấu "=" xảy ra khi x = 1

Vậy Max_B = 3/4 khi x=1

Ta có :\(A=\left|x+2\right|+\left|9-x\right|\ge\left|x+2+9-x\right|=11\)

Vậy \(A_{min}=11\) khi \(2\le x\le9\)

\(2x^2+2y^2-4xy+2x-2y+4\)

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

\(=2\left[\left(x-y\right)^2+2\left(x-y\right)\frac{1}{2}+\frac{1}{4}\right]+\frac{7}{2}\)

\(=2\left(x-y+\frac{1}{2}\right)^2+\frac{7}{2}\)

\(\Rightarrow A\ge\frac{7}{2}\)

Dấu = bn tự tính nhé