Who will help me with my sophomore math problem and how to do this sophomore math problem?

This can be solved using inequalities.

For the real numbers a, b, we always have (a-b) 2>=0, so we have a2+b 2>=2ab

So there is (a 2+z 2)+(b 2+y 2)+(c 2+z 2)=2

Therefore 2>=2ax+2by+2cz

So ax+by+cz<=1

Question 2: This question is best done by using the method of "combining numbers and shapes".

Find the intersection of the functions y=x 2 and y=x, and thus the set that satisfies y=x 2 y>x.

This set is x (-0) (1,+

There is certainly no problem in the interval (1,+, because 2 x+2 x 2 is an increment function, so log2(2 x+2 y)>2>7 8

We mainly prove that the interval x (-0) is also true.

We can change 2 x 2 x 2 to a 2 + b 2 to get it.

log2(2^x+2^y)=1-x+(-x)^(1/2)，x<0；It is evident that there is a minimum value when -x (1 2) = 1 2, i.e. when x = 1 4 there is a minimum value of 5 4.

See no. For questions like this, what is more important is the idea and the method of solving the problem.

1 question. a^2+x^2>=2ax

b^2+y^2>=2by

c^2+z^2>=2cz

So the three formulas are added together.

a^2+x^2+b^2+y^2+c^2+z^2=2 >=2(ax+by+cz)

So ax+by+cz<=1

Hey, let's think clearly.

2 questions. Because y=x 2 and 02 and 2 y>2

So 2 x+2 y>4

So log2(2 x+2 y)>2>7 8

The square of A+X is greater than zero, so square A plus square X is greater than or equal to 2AX, square B + Y square is greater than or equal to 2BY, square C + Z square is greater than or equal to 2CZ, so square A + B square + C square + X square + Y square + Z square is greater than or equal to 2ax + 2BY + 2CZ conclusion is known.

The sum of the probabilities is 1, that is, the sum of the numbers in the second row is 1, and the value of x can be calculated. The second one is empty, you just look at the first row, which numbers are between 2 3 and 9 2, find the corresponding numbers in the second row, and add up the numbers in the second row.

a8=a1+7d

a13=a1+12d

3a8=5a13 is equivalent to 3(a1+7d)=5(a1+12d), and the solution is d=-2 type dust positive 39a1

The sum of the first n terms of the equal difference series is.

sn=na1+n(n-1)Bu penitential d brother royal 2

na1-n（n-1）a1/39

n²+40n）a1/39

n-20) a1 39+400a1 39 has a maximum value of s20 when n=20

1, if the president and deputy squad leaders are all participating, then these two people have been decided, and then choose 4 people from the remaining 8 people, that is, C84, the number is 70If the president and deputy squad leaders do not participate, then you can directly choose 6 from the remaining 8 people, which is C86, and the number is 28So there are 98 kinds in total.

1) The principal and deputy squad leaders participate in C (8,4) + the principal and deputy squad leaders do not participate in C (8,6).

2) The principal and deputy squad leaders participate in C(8,4)*A(5,5)*A(2,2).

A square B square C square 0,,, divided by 2ab equals the cosine of angle c, less than zero.

Running at a constant speed, the AB direction is the direction of the combined velocity, the shortest distance.

Hook 3 strands 4 chords 5,ab The distance along the direction of the water flow is, the velocity of the water direction of the ship = , where the velocity of motion produced by the hydrostatic velocity of the boat = 8km h - 2km h = 6km h

The combined velocity is along the AB direction, 1km 6min=1km, and the hydrostatic velocity has a component perpendicular to the direction of the current, so the hydrostatic velocity of the boat is (6 +6 )=6 2km h, and the direction, at an angle of 45° with the river bank, is biased towards the direction of the current.

According to the kinematic synthesis theorem. Reducing the angle between the hydrostatic velocity and the AB (upstream of the AB) can make the combined velocity arbitrarily large, so that the shortest time is not the maximum if it is not limited by the capacity of the boat.