Let's know that the power of a-Ab = 10, the power of ab-b = 5, find the power of B-A, and then find the power of a-2ab + B =?

Let's know that the power of a-Ab = 10, the power of ab-b = 5, find the power of B-A, and then find the power of a-2ab + B =?

a²-ab=10 ①
ab-b²=5 ②
①＋②
a²-ab+ab-b²=10+5
a²-b²=15
①-②
a²-ab-（ab-b²）=10-5
a²-2ab+b²=5

How to solve the equation a + B = 17, ab = 60

When a + B = 17, deformation: a = 17-b,
Substituting (17-b) B = 60
b²-17b+60=0,
（b-12）（b-5)=0,
∴b1=12,a1=5,
b2=5,a2=12.
Solution 2: let X & sup2; + PX + q = 0,
According to Weida's theorem: a + B = - P = 17, P = - 17,
ab=q=60,∴q=60.
We obtain X & sup2; - 17x + 60 = 0
（x-12）（x-5)=0,
∴x=12,x=5,
a=12,b=5,
a=5,b=12.

Given a > 0, b > 0, a + B = 1, prove AB + 1 / AB ≥ 17 / 4
Thank you

Method 1: A, B ∈ R +, a + B = 1 ≥ 2 √ ab → 0 ≤ ab ≤ 1 / 4, let AB = x, then y = AB + 1 / AB = x + 1 / X. for the function f (x) = y = x + 1 / x, f (x) increases monotonically on (- ∞, - 1) or (1, + ∞), and f (x) decreases monotonically on (- 1,0) or (0,1)

Given that a * a-Ab = 17, ab-b * b = 12, what is ab?

a(a-b)=17
b(a-b)=12
(a-b)(a-b)=5
ab=40.8

(a + b) ^ 2 = 40, (a-b) ^ 2 = 60, a ^ 2 + B ^ 2 and ab value answer quickly!

（a+b）^2=40
a^2+2ab+b^2=40.①
(a-b)^2=60
a^2-2ab+b^2=60.②
① 2 (a ^ 2 + B ^ 2) = 100
a^2+b^2=50
① (2) 4AB = - 20
ab=-5

Given (a + b) 2 = 60, (a-b) 2 = 80, find the value of a 2 + B 2 and ab

∵ (a + b) 2 = 60, (a-b) 2 = 80, ∵ A2 + B2 + 2Ab = 60, A2 + b2-2ab = 80, ②, ∵ ① + ②: 2 (A2 + B2) = 140, A2 + B2 = 70, ∵ 70 + 2Ab = 60, ab = - 5

(a + b) quadratic = 7, (a-b) quadratic = 4. (1) find the value of a + B: (2) find the value of ab

New operation: for any number AB, there is a * b = B quadratic + 1, for example, 7 * 4 is equal to 4 quadratic plus 1 = 17, find 5 * 3 * 2

5*3
That is, a = 5, B = 3
Substituting
5*3=3²+1=10
So 5 * 3 * 2 = 10 * 2
Then a = 10, B = 2
So, sup2 + 2 = 5

On the quadratic power of AB (a-b) (a + b) a + the quadratic power of B
Column: 1. (a + b) quadratic = (a-b) quadratic + 4AB
2. The power of (a-b) = (a + b) - 4AB
It's a formula. You can add, subtract, multiply and divide as you like

（1/2+1/3+...+1/2002）（1+1/2+1/3+..+1/2001）-（1+1/2+1/3+...+1/2002）（1/2+1/3+...1/2001）

Let x = (1 / 2 + 1 / 3 +... + 1 / 2002); y = (1 / 2 + 1 / 3 +... + 1 / 2001)
The original is equivalent to
x(1+y)-(1+x)y
=x-y
=(1/2+1/3+...+1/2002)-(1/2+1/3+..+1/2001)
=1/2002