-
Anonymous users2024-02-09Root number B - root number A +8 > A - root number ab + "root number B B - root number ab - root number a a + root number ab> divided by 8 root number ab root number a + root number b = ( b - a + 8) a ( a - b) + [ 8 ( b - a) - 8 ( a + b)]*a+ b) 8 ab = ( b - a + 8) a ( a - b) + 8 a ( b- a ( b- a) (
-
Anonymous users2024-02-08a 2 + b 2-c 2) 2-4a 2b 2 flat match chaotic variance.
a 2+2ab+b 2-c 2)(a 2-2ab+b 2-c 2)[(a+b) 2-c 2][(a-b) 2-c 2](a+b+c)(a+b-c)(a-b+c)(a-b-c) The sum of the two sides of the triangle is greater than the first argument and the three carrying sides.
So a+b-c>0
a-b+c>0
a-b-c0
Three positive and one negative, multiplied less than.
So(a2+b2-c2) 2-4a2b2
-
Anonymous users2024-02-07Suppose a>b compares (a 2-ab+b 2) and a, i.e., their square comparison a 2-ab+b 2-a 2=b(b-a)< 0 compares (a 2-ab+b 2) and b, i.e., their square comparison a 2-ab+b 2-b 2=a(a-b)>0 so a> (a 2-ab+b 2)>b according to the cosine theorem cosq= [a 2+b 2-(a 2-ab+b 2)] (2ab)=1 2 q = 60 degrees.
If a
-
Anonymous users2024-02-06cosc=(a2+b2-c2) 2ab=-1 2,c=120 degrees, which is the maximum angle, because there cannot be two inside angles greater than 90 degrees within a triangle.
-
Anonymous users2024-02-05> = 0m is greater than or equal to zero.
a-b)²-c²
a-b+c)(a-b-c)
a+c>b a-b+c>0
b+c>a a-b-c<0
m-c²<0
m-c is less than zero.
-
Anonymous users2024-02-04m 0 because m = (a-b)2
When a=b, the original = 0
When a=b, the original = 0
m-c is less than zero.
-
Anonymous users2024-02-03(a 2+b 2-c 2) 2-4a 2b 2 (factorization) = (a 2 + b 2-c 2 + 2ab) (a 2 + b 2-c 2-2ab).
(a+b) 2-c 2][(a-b) 2-c 2] (refactoring).
a+b+c)(a+b-c)(a-b+c)(a-b-c) Since the sum of any two sides of the triangle is greater than the third side, the first three formulas in the above equation are all greater than 0, and only the last equation is less than 0, so [a +b -c] 4a b must be a negative value.
-
Anonymous users2024-02-02by the previous step. a-c=2b(a-c)
Move the number to the right of the equal sign to the left.
Amount. a-c)+2b(a-c)=0
The common factor (a-c) is then extracted
And I got it. a-c)(1+2b)=0
Sorted out. a-c)(2b+1)=0
This step is just a variation of the equation. Hope.
-
Anonymous users2024-02-01Because a-c = -2b (a-c).
So (a-c) + 2b (a-c) = 0 (move the number to the right of the equal sign to the left) (property of equation 1).
So (a-c) (1 + 2b) = 0 (extract the common factor (a-c)) so (a-c) (2b + 1) = 0
Actually, this step is just a variation of the equation. ^_
-
Anonymous users2024-01-31Move the right side of the equation to the left, and then mention the common a-c
-
Anonymous users2024-01-301) If a, b, and qi ji c are the three high chain balance sides of the triangle, verify that a 2-2ab- c 2+b 2b, a-b+c>0
b+c>a,a-b-c
-
Anonymous users2024-01-29Solution: 3c=12, c=4, a+b=8 , a-b=2 from the first two formulas, and a=5 and -b=3 from +
a=5,b=3,c=4
-
Anonymous users2024-01-28Because a plus b plus c = 12, and a plus b = 2c, a-b = 2, a = 5, b = 3, c = 4
Solution: Because the sum of any two sides of the triangle is greater than the third side, the difference between any two sides is less than the third side, so a 0 , b 0 , c 0 >>>More
Triangle area formula.
Area = base height 2, s = ah 2 (where a is the bottom of the triangle, do hh is the height corresponding to the bottom). >>>More
The first thing to consider is the special triangle. Because (a b 0), it can be determined that a +b is the largest edge, i.e., the hypotenuse. Therefore, it can be predicted that the sum of the squares of the two right-angled sides of a right-angled triangle is equal to the square of the hypotenuse. >>>More
Here's the answer:Solution: Polar coordinates. >>>More
When the sum of the three sides of the triangle is greater than the third side, the triangle is obtuse and acute. When the sum of the three sides of a triangle satisfies the sum of the squares of the two right-angled sides equals the square of the third side, the triangle is a right-angled triangle.