thui có ngừi gửi r

Ta có: |x+\(\dfrac{2}{3}\)|≥0 với ∀ x

⇒-|x+\(\dfrac{2}{3}\)|≤0 với ∀ x

⇒2-|x+\(\dfrac{2}{3}\)|≤ 2 với ∀ x

⇒F≤ 2

Dấu ''='' xảy ra khi ⇌ |x+\(\dfrac{2}{3}\)|=0

                              ⇌x+\(\dfrac{2}{3}\)=0

                              ⇌x=-\(\dfrac{2}{3}\)

VậyF \(_{max}\)=2 khi x=-\(\dfrac{2}{3}\) ( chọn B)

a,=(2a + b - 3c).(2a + b - 3c)

=4a\(^2\)+2ab-6ac+2ab+b\(^2\)-3bc-6ac-3cb+9c\(^2\)

=4a\(^2\)+b\(^2\)+9c\(^2\)+4ab

=2\(^2\).a\(^2\)+4ab+b\(^2\)+9c\(^2\)

=(2a+b)\(^2\)+9c\(^2\)( đáng lẽ chỗ này nó phải là -9c\(^2\) nhưng t ko ra đc )

b,=(a + 2b + 3c - 4d)(a + 2b + 3c - 4d)

=a\(^2\)+2ab+3ac-4ad+2ab+4b\(^2\)+6bc-8bd+3ac+6bc+9c\(^2\)-12cd-4ad-8bd-12cd+16d\(^2\)

=a\(^2\)+4b\(^2\)+9c\(^2\)+16d\(^2\)+4ab+6ac-8ad+12bc-16bd-24cd

=(a\(^2\)+4ab+4b\(^2\))+(9c\(^2\)-24cd+16d\(^2\))+6ac-8ad+12bc-16bd

=(a+2b)\(^2\)+(3c-4d)\(^2\)+2(3ac-4ad+6bc-8bd)

=(a+2b)\(^2\)+(3c-4d)\(^2\)+2[a(3c-4d)+2b(3c-4d)]

=(a+2b)\(^2\)+(3c-4d)\(^2\)+2(a+2b)(3c-4d)

khiếp bài dài nghoằng ra ý :(

= 3²-4\(^2\)(x-1)\(^2\)

=[3-4(x-1)].[(3+4(x-1)]

=(3-4x+4)(3+4x-4)

=(7-4x)(4x-1)

xét tứ giác FGHE có: \(\widehat{F}\)+\(\widehat{G}\)+\(\widehat{E}\)+\(\widehat{EHG}\)= 360'

⇒\(\widehat{EHG}\)= 360'-\(\widehat{F}\)-\(\widehat{G}\)-\(\widehat{E}\)=360'-90'-70'-60'=140'

Ta lại có: \(\widehat{EHG}\)+x=180'( 2 góc kề bù)

⇒x=180'-\(\widehat{EHG}\)=180'-140'=40'

Vậy x=40'

đề bài là j hả man???:v

a,= a\(^2\)+2a+b\(^2\)-2b-2ab+37

=a\(^2\)-2ab+b\(^2\)+2a-2b+37

=(a-b)\(^2\)+2(a-b)+37

⇒5\(^2\)+2.5+37= 25+10+37= 72

b,= a\(^3\)+a\(^2\)-b\(^3\)+b\(^2\)+ab-3a\(^2\)b+3ab\(^2\)-3ab-95

=a\(^3\)-3a\(^2\)b+3ab\(^2\)-b\(^3\)+a\(^2\)-2ab+b\(^2\)-95

=(a-b)\(^3\)+(a-b)\(^2\)-95

⇒5\(^3\)+5\(^2\)-95= 125+25-95= 60

a, Ta có:\(\widehat{K1}\)= \(\widehat{D1}\)(= 60')

 Mà 2 góc này ở vị trí so le trong

⇒ a//b

b, ta có a//b 

⇒ \(\widehat{E1}\)=\(\widehat{F2}\) = 90'

ta có: \(\widehat{D1}\)+\(\widehat{D2}\)= 180'( 2 góc kề bù)

⇒ \(\widehat{D2}\)= 180' - \(\widehat{D1}\)= 180' - 60'=120'

Ta lại có: \(\widehat{D1}\)=\(\widehat{D3}\)( 2 góc đối đỉnh)

⇒\(\widehat{D3}\)= 60'

uầy e đọc chả hỉu j lun :(