If a minus 1 equals B minus C less than 0 If A-1 = B-C

If a minus 1 equals B minus C less than 0 If A-1 = B-C

a

How to prove that a ^ 2 + B ^ 2 + C ^ 2 is greater than or equal to ab + BC + AC

（a+b）^2+(b+c)^2+(c+a)^2>=0
It will be proved when it is unfolded

If a minus B equals B minus C equals one fifth and a plus B plus C equals one, then the value of AC plus BC plus AB is

a-b=b-c=1/5
a-c=(a-b）+（b-c)=2/5
The above formula is square a + B - 2Ab = 1 / 25
B + C - 2BC = 1 / 25
A + c-2ac = 4 / 25
2 (a + B + C) - 2 (AB + BC + AC) = 6 / 25
2-2(ab+bc+ac)=6/25
ab+bc+ac=22/25

Simple calculation of 3.6 * 0.81 + 0.36 * 1.9

3.6*0.81+0.36*1.9
=3.6*0.81+3.6*0.19
=3.6*(0.81+0.19)
=3.6*1
=3.6

By solving the equation 3x-2 = 5 + 6x, we get____ , merge the similar items____ The coefficient is changed to 1 to get X=_____

If we solve the equation 3x-2 = 5 + 6x, we can get 3x-6x = 5 + 2 by shifting the term. If we merge the similar terms, we can get - 3x = 7. If we change the coefficient to 1, we can get x = - 7 / 3

2005×20062006-2006×20052005=______ ．

2005 × 20062006-2006 × 20052005, = 2005 × 2006 × 10001-2006 × 2005 × 10001, = 0

The triangle ABC is inscribed in the circle O, and AC 〉 BC, D is the midpoint of the superior arc ACB. Proof: ad ^ 2 = AC * BC + CD ^ 2
Verification: ad ^ 2 = AC * BC + CD ^ 2

Proof: connect ad, CD, extend BC, make de ⊥ be, DF ⊥ AC
∵ AC > BC D is the midpoint of arc ACB
D on AC
∵ arc ad = arc BD arc CD = arc CD
∴∠CAD=∠CBD
AD=BD ∠BED=∠AFD=90°
∴△BDE≌△ADF(AAS)
DF=DE DC=DC
△CDE≌△DCF(HL)
CF=CE
∴AD²=AF²+DF²=AF²+CD²-CF²=(AF+CF)(AF-CF)+CD²
=AC*(BE-CE)+CD²=AC*BC+CD²
AD²=AC*BC+CD²

1,8,16,26,39,56,78,…
Given the formula, the nth term can be calculated,

There are infinitely many such formulas. Just find one to satisfy the given term, and give one as follows:
an=(n³-3n²+44n-36)/6.

As shown in the figure, in △ ABC, ad is the middle line on the edge of BC, f is a point on the edge of AD, and aeeb = 16, and the ray CF intersects AB at point E, then AFFD is equal to___ ．

As shown in the figure: through point D, make DG ‖ EC intersect AB with G, ∵ ad is the midline on the edge of BC, ∵ GD is the median line of △ BEC, ∵ BD = CD, BG = Ge. ∵ aeeb = 16, ∵ AEEG = 13 ∵ DG ‖ EC, ∵ AEEG = AFFD = 13

How much is 3.19M minus 14.5cm?

3.19M minus 14.5m
＝319-14.5
= 304.5cm
= 3.045m