How can a × 2 ^ 3 + 1 / 2B * (- 4) + 5 = 199 be reduced to 4a-b = 97

How can a × 2 ^ 3 + 1 / 2B * (- 4) + 5 = 199 be reduced to 4a-b = 97

4A ^ 2 + 4-4a + B ^ 2 + 2B = - 4 find the value of a ^ B

Find the value of a and B; a ^ 2-2b = - 1, B ^ 2-4a = - 4

Two formula addition
a^2-2b+b^2-4a=-1-4
a^2-4a+4+b^2-ab+1=0
(a-2)^2+(b-1)^2=0
The sum of two squares equals zero
Then A-2 = 0 and B-1 = 0
Then a = 2 and B = 1

When a = 0.8 and B = 0.5, the value of 4A - (2b × B-A) is obtained

4a-(2b×b-a)
=4a-2b²+a
=5a-2b²
=5×0.8-2×0.5×0,5
=4-0.5
=3.5

It is stipulated that a} B = 4a-2b, if x} (8} 7) = 24, the value of X is calculated
emergency

x◇(8◇7)=24
x◇(4×8-2×7）=24
x◇18=24
4x-2×18=24
4x=60
x=15
[analysis: this kind of problem can be worked out by drawing gourd according to the sample step by step,

a*a+b*b-4a+2b+5=0 ab=?

(x + 1) ^ + (Y-2) ^ = 0, then x + y =; if a ^ + B ^ - 4A + 2B + 5 = 0, then ab=

(x+1)^+(y-2)^=0
x+1=0
y-2=0
x=-1
y=2
x+y=-1+2=1
a^+b^-4a+2b+5=(a-2)^+(b+1)^=0
a=2
b=-1
ab=-2

If a ^ 2-4a + B ^ 2 + 2B + 5 = 0, then the value of ab

If a ^ + B ^ - 4A + 2B + 5 = 0, then ab=
I'm going to trouble you again~`

The original formula is: (A-2) ^ 2 + (B + 1) ^ 2 = 0
Because (A-2) ^ 2 and (B + 1) ^ 2 are both > 0
So a = 2, B = - 1
So AB = - 2

Given a + B + 4a-2b + 5 = 0. Find the value of a + B / 2-ab

A = - 2, B = 1, a + B / 2-AB = 4 + 1 / 2 + 2 = 13 / 2, you will judge well!