Calculation: the square of A-16 parts of a-5a + 6 divided by the square of a + 5A + 4 parts of A-4 × A-3 parts of A-4

Calculation: the square of A-16 parts of a-5a + 6 divided by the square of a + 5A + 4 parts of A-4 × A-3 parts of A-4

[(a²-5a+6)/(a²-16)]÷[(a²-4)/(a²+5a+4)]×[(a-4)/(a-3)]
=(a²-5a+6)(a²+5a+4)(a-4)/[(a²-16)(a²-4)(a-3)]
=(a-2)(a-3)(a+1)(a+4)(a-4)/[(a+4)(a-4)(a+2)(a-2)(a-3)]
=(a+1)/(a+2)
=(a+2-1)/(a+2)
=1 -1/(a+2)

Given the square of a + 5A + 1 = 0, try to find the value of the following scheme,
1) A + a 1 / 2
2) A square + a one in two
3) A cube + a cube
4) A square + a quarter

Divide the original formula by a to get a + 5 + a 1 / 2 = 0, so a + a 1 / 2 = - 5, square a, square a + 2 + a 1 / 2 = 25, so a square + a 1 / 2 = 23, square a, square a + 2 + a 1 / 4 = 529, square a + a 1 / 4 = 527, then square a + a 1 / 2 times a + a 1 / 2 to get a cube + A + a 1 / 1 + a 1 / 3 = - 105, Then cube a + cube a = - 105 + 5 = - 100
So the result is: - 5,23, - 100527

If a * a + 3a-1 = 0, then a * a * a + 5A * a + 5A + 18 =? The square cube can be abbreviated

That is a & sup2; + 3A = 1
The original formula is a & sup3; + 3A & sup2; + 2A & sup2; + 5A + 18
=a(a²+3a)+2a²+5a+18
=a×1+2a²+5a+18
=2a²+6a+18
=2(a²+3a)+18
=2×1+18
=20

Given the square of a-5a + 1 = 0 (a ≠ 0), (1) find the value of the square of a + the square of a / 1, (2) find the value of A-1 / A

The square of a - 5A + 1 = 0;
a-5+1/a=0;
a+1/a=5;
(1)a²+1/a²=(a+1/a)²-2=25-2=23;
(2)(a-1/a)²=a²+1/a²-2=23-2=21;
∴a-1/a=±√21;

What's the product of 2 / 3 plus 1 / 5 divided by 3 / 4?

(2 / 3 + 1 / 5 ÷ 3 / 4) what's 4 / 4

There are four numbers ABCD (ABCD is not zero) which satisfy the equation a * half = b * two thirds = C * four fifths = D. please change the four numbers ABCD from small to large
In order

d＜c＜b＜a

If ABCD is four numbers that are not equal to 0 and a times 3 / 2 = B times 3 / 3 = C times 4 / 5 = D, then which number is the largest and which number is the smallest among the four numbers ABCD

If it's a choice, judgment type, you can directly take the special case a = 2, then B = 3, C = 15 / 4 > 3, d = 3, then C is the largest, a is the smallest. If it's a solution problem, you can assume a = 2x (why not set it as X, because 2x and X are also unknowns, and 2x is more convenient than x in the process of calculation), then B = D = 3x, C =

If the product of four numbers ABCD = 25, and a is not equal to B, C is not equal to D, then a + B + C + D =?

Let's limit a, B, C and D to integers, not too many
1×(-1)×(-5)×5=25
So a + B + C + D = 0

There are four numbers ABCD (ABCD is not zero) which satisfy the equation a by half = B by two thirds = C by four fifths = D
Arrange ABCD from small to large
Primary school level···

1/2a=2/3b=4/5c=d
So a: B: C: D = 8:6:5:4
So a > b > C > D

There are four numbers ABCD (ABCD is not zero), which satisfy the equation that a multiplied by half equals B multiplied by two thirds equals C multiplied by five quarters equals D

a/2=2b/3=5c/4=d
So a = 2D, B = 3D / 2, C = 4D / 5
D can be any number, such as d = 1, then a = 2, B = 3 / 2, C = 4 / 5