If half of the equation (x-1) = 5 and one third of the equation (ax-4) = 6, a=

If half of the equation (x-1) = 5 and one third of the equation (ax-4) = 6, a=

1/2 (x-1)=5
1/3(ax-4)=6
----------
x-1=10
ax-4=18
--------
x=11
ax=18+4=22
--------------
a*11=22
a=2

125 / 10 * 8 calculated by simple method

125÷10×8
=125×8÷10
=1000÷10
=100
or
125÷10×8
=125÷5÷2×8
=（125÷5）×（8÷2）
=25×4
=100

If a > 0 and b > 0 are constants, then the limit Lim [(^ x + B ^ x) / 2] ^ 3 / x =?
Calculation process
lim[(a^x+b^x)/2]^3/x=？
Missed ~ ~ sorry

a^3/2 * b^3/2
exp{lim 3ln(a^x+b^x)/2 /x}
Luo
=exp{lim 3*(lna*a^x+lnb*b^x)/(a^x+b^x)}
=exp{(3lna+3lnb)/2}
=a^3/2*b^3/2
Use the law of lobita and it's out
I got it right. Give me points, landlord~

If the expansion of polynomial (MX + 4) (2-3x) does not contain x term, then M=______ ．

∵ (MX + 4) (2-3x) = 2mx-3mx2 + 8-12x = - 3mx2 + (2m-12) x + 8 ∵ after expansion, there is no X term ∵ 2m-12 = 0, that is, M = 6

Simple calculation of fraction multiplication in sixth grade volume I
9/5-16/7*9/5

Is it five out of nine?
5/9-7/16*5/9
=5/9*（1-7/16）
=5/9*9/16
=5/16

LNX integral and (LNX) square integral, where x belongs to 1 to e, find the size of these two integrals

Answer: ∫ (1 to e) lnxdx = (xlnx-x) | (1 to e) = elne-e-ln1 + 1 = 1 ∫ (1 to e) (LNX) & sup2; DX = [x (LNX) & sup2; - 2xlnx + 2x] | (1 to e) = e (lne) & sup2; - 2elne + 2E - (ln1) & sup2; + 2ln1-2 = E-2 ∫ (1 to e) (LNX) & sup2; DX

When at least one root of the equation x ^ 2 + ax + 2 = 0 is less than - 1, find the value range of real number a (find the specific process)

1. Discuss whether the equation has two roots. When the discriminant is 0, a = positive and negative double root sign 2, then a = double root sign 2 can hold
2. When the equation has two solutions
First, a is greater than double root 2 or a is less than negative double root 2
And because f (0) = 2
So both roots are on the negative half axis of X
There are two kinds of feelings
In the first case, if both roots are less than - 1, the axis of symmetry is less than - 1 and f (- 1) > 0
The second is that if a root is less than - 1, then f (- 1) 0
I haven't worked on this topic for a long time. Maybe some discussions are unnecessary

Calculated by vertical formula. 83.4 △ 39.84 △ 24.95 △ 5

lim(n->∞)[√(1+cosπ/n)+√（1+cos2π/n）+…… +√（1+cosnπ/n）]*1/n=

Transform to integral = ∫ (from 0 to 1) √ (1 + cos π x) DX = ∫ (from 0 to 1) √ [cos & # 178; (π X / 2)] DX = ∫ (from 0 to 1) | cos (π X / 2) | DX = ∫ (from 0 to 1) cos (π X / 2) DX = (2 / π) ∫ (from 0 to 1) cos (π X / 2) d (π X / 2) = (2 / π) sin (π X / 2) | (from 0 to 1) = (2 / π) ∫ (from 0 to 1) = (2 / π) ∫ (from 0 to 1) ∫ (π X / 2) ∫ (π X / 2) ∫ (from 0 to 1) = (

Using image method to solve equations X-Y = 3,2x + y = 6

x=3,
X-Y = 3 can be changed into a function y = x-3
2X + y = 6 can be changed into function y = 6-2x
Draw the image at a point, the abscissa is 3, the ordinate is 0
So x = 3