-(a squared b) squared * (a squared b) cubic calculation,

-(a squared b) squared * (a squared b) cubic calculation,

-The second power of (a) the second power of (b) * (a) the third power of (b)
=-(2 + 3) power of (a)
=-The second power of (a) and the fifth power of (b)
=-(2 × 5) power of a and 5 power of B
=-A to the 10th power B to the 5th power

Square of (a-b) * (B-A) cube + ((a-b) square)

=-(a-b)^5+(a-b)^4
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Cube of 2 (a-b) - square of 4 (B-A) =?

=2(a-b)³-4(a-b)²
=2(a-b)²[(a-b)-2]
=2(a-b)²(a-b-2)

If the square of a = 4 and the cube of B = - 3, then a / b=

If the square of a = 4,
a=±2；
Cube of B = - 27,
b=-3
Then a / b = ± 2 / 3;
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The cube of - 2 (a-b) + the square of 4 (a-b) - (B-A) + the cube of (a-b) is merged into the same category

-2(a-b)+4(a-b)-(b-a)+(a-b) = -2(a-b)+(a-b)+4(a-b)-(a-b) =-(a-b)+3(a-b) =(a-b)[-(a-b)+3] =(a-b)(b-a+3)

It is known that a and B are opposite to each other, C and D are reciprocal to each other, the absolute value of M is equal to 2, and the value of | a + B | / 2m to the second power + 4m-3cd is obtained

That is, a + B = 0
cd=1
m=±2
So M & # 178; = 4
So the original formula = | 0 | / (2 × 4) + 4m-3 × 1
=0+4m-3
=4 × (- 2) - 3 = - 11 or = 4 × 2-3 = 5

Given that a and 1 / (a ^ 2-2) are reciprocal, how many real numbers satisfy the condition?

a×1／（a²-2)=1
a=a²-2
a²-a-2=0
(a-2)(a+1)=0
A = 2 or a = - 1

a. And a / A2_ 1 is reciprocal to each other and satisfies the condition number of real number a

a×a/(a²-1)=1
a²=a²-1
0=-1
unsolvable
No a satisfies the condition

What is the reciprocal of the negative real number a?

1/a

For real numbers a, B, "B (B-A)"

If B = 0, then B (B-A) = 1
Then multiply B by 178 on both sides;
ab>=b²
b²-ab