\(x^2-2x-1=\left(x^2-2x+1\right)-2=\left(x-1\right)^2-2\ge2\)

Dấu "=" xảy ra khi x = 1

\(x^2+3x+7=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu "=" xảy ra khi x = - 1,5

Trời thêm tuổi mới năm thêm thọ

xuân khắp đất trời phúc lộc khắp nhà

This is math, not Vietnamese

Bài 1:

\(\Delta ABD-\text{vuông}-\text{tại}-A\)

\(\Rightarrow\tan B=\dfrac{AD}{AB}\)

\(\Rightarrow AD=\tan50^0\times AB\approx2,68621\left(cm\right)\)

\(S_{ABCD}=\dfrac{1}{2}\times\left(AB+CD\right)\times AD\)

\(\Rightarrow CD=\dfrac{2S_{ABCD}}{AD}-AB\approx5,15569\left(cm\right)\)

Kẻ BH _I_ CD.

\(ABHD-\text{là}-h.c.n.\left(\widehat{BAD}=\widehat{ADH}=\widehat{DHB}=90^0\right)\)

\(\Rightarrow AB=DH=2,254\left(cm\right)-\text{và}-AD=BH\approx2,68621\left(cm\right)\)

\(\Rightarrow HC=DC-DH\approx2,90169\left(cm\right)\)

\(\Delta HBC-\text{vuông}-\text{tại}-H\)

\(\Rightarrow BC^2=BH^2+HC^2\left(ptg\right)\)

\(\Rightarrow BC=\sqrt{BH^2+HC^2}\approx3,95418\left(cm\right)\)

~ ~ ~

\(\Delta HBC-\text{vuông}-\text{tại}-H\)

\(\Rightarrow\tan\widehat{HBC}=\dfrac{HC}{HB}\)

\(\Rightarrow\widehat{HBC}\approx47^012'29,89"\)

\(\Rightarrow\widehat{ABC}=\widehat{ABH}+\widehat{HBC}\approx137^012'29,89"\)

mà \(\widehat{ABC}+\widehat{BCD}=180^0\) (2 góc trong cùng phía, AB // CD)

\(\Rightarrow\widehat{BCD}\approx42^047'30,11"\)

\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

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

\(=1-3ab+3ab\left(a+b\right)^2\)

= 1

 a) 136.(-47)+36.47

= -136.47 + 36. 47

=47 . (-136 + 36 )

=47 . (-100)

=-4700

b) (-48).72+36

=(-48).2.36+36

= -96.36+36

=(-96+1).36

= -95.36=-3420

Số sách ở ngăn thứ nhất là :

( 85 - 13 ) : 2 = 36 ( quyển )

Đáp số : 36 quyển sách

\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)

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

\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)

\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)