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A (1, - 1), B (- 2, - 7), C (0,3) are collinear
You have the wrong title
AB=(-2-1,-7+1)=(-3,-6)
BC=(0+2,3+7)=(2,10)
It doesn't matter. You can't draw
When 3A square + ab-2b square = 0 (a, B are not 0), find the value of A-A part of b-ab part of a square + b square
3a²+ab-2b²=(3a-2b)(a+b)=0
3a-2b = 0 or a + B = 0
So B = 3A / 2 or B = - A
b=3a/2
Original formula = 2 / 3-3 / 2 - (1 + 9 / 4) / (3 / 2) = - 3
b=-a
The original formula = - 1 + 1 - (1 + 1) / (- 1) = - 2
Find 3A square + ab-2b square = 0 and a, B are not 0. Find the value of a / B-B / a - (a square + b square) / ab
If the positive integer solution of the inequality - K-X + 6 > 0 about X is 1.2.3, there is one step I don't understand
In the middle, there is a step 3 〈 - K + 6 〈 = 4. How did it come about?
The inequality can be reduced to - K + 6 > X
If x is 1 2 3
Side-k + 6 is between three and four
If - K + 6 is 3
X cannot take three
If - K + 6 is 4
X can take 1, 2, 3
If - K + 6 is greater than 4
X can take 1, 2, 3, 4
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