Given that a = the third power of X-5 times the second power of X, B = the second power of x-11x + 6, find (1) a + 2B; (2) when x = - 1, find the value of a + 5B

Given that a = the third power of X-5 times the second power of X, B = the second power of x-11x + 6, find (1) a + 2B; (2) when x = - 1, find the value of a + 5B

A+2B
=(x³-5x²)+2(x²-11x+6)
=x³-5x²+2x²-22x+12
=x³-3x²-22x+12
x=-1
A+2B
=x³-3x²-22x+12
=-1-3+22+12
=30
Find the range of function y = log small 3 x + log small X 3-1
RT
Finding the range of function y = log3 (x) + logx (3) - 1
When x > 1, log3 (x) > 0, logx (3) > 0,
log3 (x)+logx (3)≥2√(log3 (x)•logx (3))=2.
y≥1.
Two
It is known that the 2 / 3 power of (- radical a) is 5, then a =?
2 / 3 power of (- radical a) = 5
=[(-√a)^2]^(1/3)=a^(1/3)=5
therefore
a=5^3=125
The value range of the function y = log with 1 / 2 as the base [(x ^ 2) + 6x + 17] is
Y = log is based on 1 / 2 [(x ^ 2) + 6x + 17]
Let u = (x ^ 2) + 6x + 17 = (x + 3) ^ 2 + 8
The range of easily obtained u is [8, positive infinity]
Y = log based on 1 / 2 u > = 8
The bottom of log is 1 / 2, 8 = - 3
Therefore, according to the image, we can get the value range as (negative infinity, - 3)
Remember to add points!
In this paper, we change the result of (3A's square B's negative third power) negative first power into a form with only positive integral exponential power
b^3/(3a^2)
The monotone increasing interval of log half (x ^ 2 + 4x-12) is?
There should be a process, thank you
Y = Log1 / 2 (x ^ 2 + 4x-12) the domain is: (2, positive infinity) and (negative infinity, - 6)
Y = Log1 / 2 (x ^ 2 + 4x-12) is composed of y = Log1 / 2 (T) and T = x ^ 2 + 4x-12. Y = Log1 / 2 (T) is a decreasing function on the domain,
T = x ^ 2 + 4x-12 is an increasing function on (2, positive infinity) and a decreasing function on (negative infinity, - 6)
Therefore, y = Log1 / 2 (x ^ 2 + 4x-12) is a monotone decreasing function on (2, positive infinity) and an increasing function on (negative infinity, - 6)
If the polynomial 2mx ^ 3 + 3nxy ^ 2-2x ^ 3-xy ^ 2 + y contains no cubic term, then 2m + 3N =?
There is no cubic term in 2mx ^ 3 + 3nxy ^ 2-2x ^ 3-xy ^ 2 + y
Therefore, it does not include the four items 2mx ^ 3 + 3nxy ^ 2-2x ^ 3-xy ^ 2
2mx^3+3nxy^2-2x^3-xy^2
=（2m-2）x^3+(3n-1)xy^2
=0
So 2m-2 = 0, 3n-1 = 0
So m = 1, n = 1 / 3
2m+3n=2+1=3
The original formula = (2m-2) x & # 179; + (3n-1) XY & # 178; + y
If there is no cubic term, the coefficients of the first two terms are 0
2m-2=0
3n-1=0
So 2m = 2, 3N = 1
So yuan = 2 + 1 = 3
The monotone increasing interval of function f (x) = log12 (2x2-5x + 3) is______ .
Let g (x) = 2x2-5x + 3, then G (x) is a decreasing function when x < 1, and G (x) is an increasing function when x > 32. And y = log12u is a decreasing function, so f (x) = log12 (2x2 − 5x + 3) is an increasing function when (- ∞, 1). So the answer is: (- ∞, 1)
It is known that the polynomial MX ^ 2 + 3nxy ^ 2-2x ^ 3-xy ^ 2 + 2x + y about X and Y does not contain cubic term, so we can find the value of 2m + 3N
Because it does not contain three times, then
mx^3+3nxy^2-2x^3-xy^2+2x+y
=(m-2)x^3+(3n-1)xy^2+2x+y
m-2=0 3n-1=0
m=2 n=1/3
2m+3n=2*2+3*1/3=4+1=5
You are also from Xinghai. It seems that the title of this problem is wrong. I think we should change the square into a cube, and then I'll calculate it. That's right
Increasing range of interval function (2x-2)
If the base of y = log2 (x2-2x-3) is greater than 0, it is an increasing function. Therefore, the increase or decrease of y = log2 (x2-2x-3) is determined by the true number. F (x) = x & # 178; - 2x-3 = (x-1) &# 178; - 4 when x > 1, f (x) is an increasing function