Mean inequality: let a be greater than or equal to 0, b greater than or equal to 0, a + B / 2 = 1 a times the maximum value of 1 + B under the root sign

Mean inequality: let a be greater than or equal to 0, b greater than or equal to 0, a + B / 2 = 1 a times the maximum value of 1 + B under the root sign

a²+b²/2=1 , 2a²+b²=2
a√(1+b²) =√[a² (1+b²)]=√1/2[2a² (1+b²)]
=√2/2•√[2a² (1+b²)]
≤√2/2•[(2a²+ (1+b²))/2]
=√ 2 / 2 · (3 / 2) = 3 √ 2 / 4

If a and B are rational numbers and a + B radical 3 = (5-2 radical 3) square, then what are a and B equal to

(5-2 root number 3) square = 25-20 root number 3 + 12 = 37-20 root number 3
a=37 b=-20

The algorithm for defining the operation "@" is as follows: x@y= X + y − 4, then（ 2@6 ）@8=______ .

According to the meaning of the title: 2@6=
2+6−4=
4=2，
Then（ 2@6 ）@ 8=2@8=
2+8−4=
6．
So the answer is:
Six

The operation rule of defining operation ＊ is: X ＊ y = radical XY + 4, then 4 ＊ 8=

4※8=√﹙4×8+4）
=√36
=6

If the operation of "@" is defined as: x@y= The root sign X-1 is the root sign x + 1（ 2@3 @8 Please analyze it carefully

Radical 2 + 1

As shown in the figure, in the triangle ABC, ab = 4, AC = 3, D is the midpoint of the edge BC. Calculate the vector ad multiplied by the vector BC=

bc=ac-ab
ad=ab+1/2bc=1/2(ab+ac)
ad.bc=1/2 (ab+ac)(ac-ab)
=1/2[ ac.ac - ab.ab ]
=1/2[3*3-4*4]
=-7/2

The results show that AD and CE are two heights of △ ABC, (1) prove that △ BDE ∽ △ BAC; (2) if AC = 10,5bd = 3ba, find the length of de

1. AD and CE are two heights of △ ABC,
A. C, D, e four points are in the same circle (the distance from deac to AC end point is equal)
Angle BDE = BAC, B = b
△BDE∽△BAC
2.:△BDE∽△BAC
BD/BA=DE/AC
DE=0.6*10=6

It is known that ad is the midline of △ ABC, e is a point on AC, and be is connected to ad to F, and AE = EF Hurry up. If you play well, you can get extra points

Conclusion: extend ad to G so that DG = ad, connect BG ∵ DG = ad; BD = DC ∵ BDG = ≌ △ ADC (SAS) &

In the triangle ABC, ab = AC, ∠ a = 90 °, CD bisection ∠ ACB, e is on AC, and AE = ad, EF ⊥ CD intersects BC in F. it is proved that BF = 2ad

DG ⊥ BC is made by D, and EF cross CD is h
∵△CHE△≌CHF（ASA）
∴CE=CF
∵△CDA△≌CDG（AAS)
∴ CA=CG,AD=DG
∴EA=FG
And ∵ AC = AB, ∵ a = 90 degrees
The △ DGB is an isosceles right angle △
∴DG=GB
Because AE = ad
So EA = FG = BG = ad and BF = BG + GF
So BF = 2ad

As shown in the figure, ad and CE are the height of △ ABC, BC = 12, ab = 10, ad = 6, and find the length of CE

∵ AD and CE are the high values of ∵ ABC,
∴S△ABC=1
2BC•AD=1
2AB•CE，
∴1
2×12×6=1
2×10×CE，
CE = 7.2