The first year of high school topic is better to give an explanation, thank you!

A is wrong, this one only pays attention to the numerical value, and does not notice the unit of Avogadro's constant, which is in mol -1, which stands for "per mole". And it may also be other Cs, such as C13, C14

b is wrong, the unit is as above.

c is false, e.g. 1molH2O is Avogadro's constant for H2O molecules, but if you find H atoms, it is twice Avogadro's constant.

d, it should be 2g

e, the first half of the sentence is correct, and the second half of the sentence is inaccurate. It should be: a mole is a unit of quantity that represents a substance, and each mole of a substance contains so many particles that make up the substance that have the Avogadro constant.

That is, substances are not necessarily made up of molecules, such as the metal Fe, which is composed of atoms, so each mole of Fe contains one Fe atom).

f, unit error.

g, h, no, these are two different concepts. The molar mass, when measured in "g mol", is numerically equal to the relative molecular mass.

i, without units, the molar mass of sulfate is 96g per mol,

e, g are correct. A should be C-12 instead of carbon, and carbon also has isotopes, such as C-14, etc.; In b, Avogadro's constant has a unit, and its unit is mol; The amount of substance that should be in c is 1mol; Isotopes such as deuterium, deuterium, and tritium are also not considered; The unit of mass in f is kg or g, not g·mol 1;h is the same as a and d; The molar mass of sulfate in i should be 96g mol.

Take L1 as an example:

Because l1: 3x+4y-12=0 is parallel to the straight line 3x+4y=0, the straight line 3x+4y=0 passes the origin, and any of the above points is taken, such as m=(-4.).+3), calculate the angle between the straight line 3x+4y=0 and the x-axis.

In the same way, the angle between L2 and the X-axis is calculated.

They subtracted each other to know that this round Hui Sen replied about the angle between L1 and L2.

|x|>=1, i.e., x<=-1, x>=1

then f(x)>=1

The f[g(x)] range is x>=0

0<=x<1 is also missing

then |x|<1

There should be 0<=-x<1

So -1=1 and -1=1.

f(x)>=0

So here the g(x) range is contained in x<=0, x>=1 is a quadratic function, then the range is <=a number or 》=a number.

So the range of g(x) is g(x)<=0

The range of g(x) is all possible x values at f(x) [0, positive infinity).x|>=1 is clearly true.

x|< 1 f(x)>0 -1< x<0, so the range of g(x) is (- 0) (1,+ < p>

x + (a-1) x + 1 0 is a false proposition, which means that x + (a-1) x + 1 > 0 is constant, and the minimum value is greater than 0 or x + (a-1) x + 1 = 0 is unsolved, and b -4ac<0 can be found.

then (a-1) 2-4<0

2<(a-1)<2

then, a (-1,3).

Solution 1: If f(x)=f(1-x) is satisfied, then b=-1 can guarantee its validity.

i.e. f(x)=x2-x

Solution: 1 is known from f(x)=f(1-x) x 2+bx=(1-x) 2+b(1-x) to reduce the disadvantage cover to 2(b+1)x=b+1 for any x.

b=-1 so f(x)=x 2-x ;

2 From 1, we can see that this inequality is (a is the bottom, omitted and not written, you write it when you answer the question) 1+log(x 2-x)>log(2x-2).

So log[(2x-2) (x 2-x )]1 then 0<(2x-2) (x 2-x )1, when a 2 x> 2 a;

When a 2 is balanced x>1;

Let's take the college entrance examination, go and communicate with the teacher,,, the best helper of the teacher's number,,, question 7, draw a feasible domain, .In fact, the optimal solution (1,3) is that y=ax+z passes the fixed point (1,3) and a is its slope, so that a is greater than the slope of x+y-4=0 1, that is, y=ax+z will not have a second intersection with the feasible domain.

Question 10: y 1-x

ax+by ax+b(1-x)=(a-b)x+bax+by=<1 can be equivalent to (a-b)x+b 1 when a b.

a-b)x+b at x=1 to achieve the maximum value a

Therefore, the value range of a and b is as follows:

0≤b≤a≤1

When a(a-b)x+b achieves the maximum value b at x=0

Therefore, the value range of a and b is as follows:

0≤a≤b≤1

In summary. 0≤b≤1,0≤a≤1

So the area is 1

Solution: Factoring 3a 2 + ab-2b 2 yields:

3a 2+ab-2b 2=(a+b)(3a-2b)=0 so there is: a+b=0, or 3a-2b=0

Because a≠0 and b≠0, b a=-1 or 3 2 ---

a/b - b/a - a^2+b^2）/ab=[a^2-b^2-(a^2+b^2)]/(ab)=-2b^2/(ab)

2b a is derived from the conclusion of the formula, which can be obtained:

a b - b a - a 2+b 2) ab = -2b a = 2 or -3

Since a>0 and the symmetry line is x=a 4 is always on the right side of the y-axis, so to make it always greater than or equal to 0 at (0, positive infinity), as long as <=0, 0<=a<=8, then the probability is 8 10 = 4 5