Knowing ab a b 1 3, bc b c 1 4, ac a c 1 5, find the value of abc ab bc ca.

Yes. ab+bc+ac abc = 1 a + 1 b + 1 c = 6 so abc (ab + bc + ac) = 1 6

abc (ab+bc+ac) = a+b+c, this is unequal.

The numerator can be divided, but the denominator cannot be divided in this way.

The ab (a+b) above you is upside down as (a+b) ab=1 a+1 b, and you didn't say ab (a+b)=a+b.

High difficulty! I didn't even read! I don't understand!

Amakusa can be peeled by hand, but the inside is the same as an orange? Is it orange or orange?

Native to Japan, its main characteristics: the tree is medium, the fruit is oblate and spherical, the weight of a single fruit is about 200 grams, the size is neat, the skin is light orange, thin, the fragrance is fresh, the flesh is orange, soft and juicy. Unisexual fruiting is strong, generally seedless, the fruit ripens at the end of December, and can be stored for 4 months.

Amakusa has a high yield and high quality, and will be a promising hybrid citrus variety in our county in the future.

In 1982, the Kuchinotsu branch of the Fruit Tree Experimental Field of the Ministry of Agriculture and Water of Japan was formed by crossing "Page" orange with "Kiyomi" and "Kotsu Zaosheng" as intermediate female parents. After many years of variety adaptability and hereditary identification tests, it was named "Amakusa" orange in 1993 and registered as "orange orange No. 5". The tree is medium-sized, the canopy expands more slowly, and the young trees are erect and gradually open after fruiting.

The density of the branch tips is medium and dense, like "clear sight", clumpy, denser than ordinary Wenzhou mandarins; The leaf size is medium, slightly smaller than that of ordinary Wenzhou mandarins, and larger than that of "Page" oranges. Fruit: pollen, fertile, but self-sterile, parthenocarpic, generally seedless; If there is a mixture of other varieties, there are more than 10 seeds.

The weight of a single fruit is about 200 grams, and the size is neat. The fruit type is oblate and spherical, and the fruit type index is left and right. The peel is pale orange, the same as the "Page" orange, and the coloring is early, in mid-December.

The pericarp is thin, about 3 mm in the equatorial part. It is slightly more difficult to peel, but it is easier than "Kiyomi". The surface of the fruit is as smooth as that of "Page" oranges, the oil bag is large and thin, and the peel has the smell of "Krimadin" red oranges, and there is also a good aroma of sweet oranges.

The flesh is orange, lighter than Wenzhou mandarin, and the same as "Kiyomi" and "Peppa" oranges. The meat is soft and juicy, and the juice is harder than Kiyomi and softer than Page. The walls of the capsule are thin and have no bitter taste.

The sugar content of the ripening fruit juice is 11-12%, and the acid is about 1%. The rate of acid reduction varies greatly depending on the cultivation area. Ripening period: late December, late January of the following year, excellent quality and good flavor.

From the known: a = b + 3 5 c = b-3 5 substituted into a + b + c = 1, 3b = 7 25

ab+bc+ac=(b+3 5)*b+(b-3 5)*b+(b+3 5)*(b 3 5)=3b -9 25=7 25 9 25= 2 25 Hope it helps, thank you!

∵ a²+b²+c²=1

a²+b²+c²-（ab+bc+ca）=1-（ab+bc+ca）

2[a²+b²+c²-（ab+bc+ca]=2[1-（ab+bc+ca)]

2a²+2b²+2c²-2ab-2bc-2ac=2-2(ab+bc+ca)

a²-2ab+b²)+b²-2bc+c²)+a²-2ac+c²)=2-2(ab+bc+ca)

a-b)²+b-c)²+a-c)²=2-2(ab+bc+ca)

a-b=b-c=3/5

a-b)+(b-c)=3/5+3/5

a-c=6/5

a-b)²+b-c)²+a-c)²

54/25=2-2(ab+bc+ca)

ab+bc+ca)=-4/50

Your satisfaction is my greatest joy!

a+b+c) is early * (a+b+c) eggplant = a +b difference +c +2 (ab + bc + ac) = 16

There is a +b +c =8

(a+b+c) square = the square of a + the square of b + the square of c + 2ab + 2bc + 2ca 11 can be solved directly with this formula

Prompt you with two ways: one (a+b+c) 3 violence; Second, a 3+b 3+c 3 is expressed in those in the question, in fact, there are many ways to do this, and the fastest can be done by using the recursive form of the number series a n=a 3+b 3+c 3.

a-b=5,b-c=3

The addition of the two formulas yields a-c=8

2(a²+b²+c²-ab-bc-ac)

2a²+2b²+2c²-2ab-2bc-2ac=(a-b)²+b-c)²+a-c)²=25+9+64=98

a²+b²+c²-ab-bc-ac=49

Good luck.

Since a+b+c=0 , a 2+b 2+c 2=1 we can find ab+bc+ac=-1 2 from (a+b+c) 2=0

So (ab+bc+ac) 2=1 4 gives a b +b c +c a =1 4

So from (a + b +c ) 2=1 we get a 4 + b +c 4 = 1 2