Senior One Mathematics Explain in detail, Senior One Mathematics Explain in detail

Uh, first of all, ah, upstairs did something wrong.

Besides. I can't learn well, maybe the method is not very good

But I think it should be right.

As upstairs, the perpendicular line is made from a and b to the x-axis, and the vertical foot is d, e

The coordinates of point d (3 5,0) ao=bo=oc=1 should be calculated.

So, I know that sin b=sin (2 3 -a)=sin2 3 ·cosa - cos 2 3 ·sina

sin b=be ob=be 1 So the ordinate of point b can be known, and the abscissa is obtained by the Pythagorean theorem.

I won't let you count the time relationship...Forgive me

Do you get the ?..

Let aoc= , then sin = 4 5 , cos = 3 5 , let d be the intersection of the circle o and the y axis of the positive semi-axis, aob is a regular triangle, bod = 30°, sin( 30°).

sinα·cos30°－cosα·sin30°cos(α-30°)

cos·cos30° sin ·sin30°b point coordinates [(4 3 3) 10 ,(3 3 4) 10].

b（-4/5，3/5）

Because the triangle ABC is a regular triangle, it is not difficult to conclude that the radius of the circle is 1, which is made as AD and BE perpendicular to the X-axis, and it is not difficult to prove that the triangle OAD and the triangle OBE are congruent, so OE = AD = 4 5, because the point E is in the opposite direction of the X axis, so the abscissa is -4 5, so it is ......

Answer. What does point c mean, and is there a missing item?

(1) Let x(x,y), and use the vector op vector ox on op to obtain x=2yxa=(1-x),xb=(5-x,1-y)xa·xb=(x-1)(x-5)+(y-1)(y-7)=5y 2-20y+12=5(y-2) 2-8

For ox(4,2), the minimum value of xa·xb is -8(2)xa=(-3,5) xb=(1,-1)cos axb=(xa·xb) (|).xa|·|xb|)=-4√17/17

Let vector a end point a, vector b end point b, vector c end point c.

If we can prove that the vector ab vector ac can prove that a, b, and c are collinear, that is, the end points of vectors a, b, and c are in a straight line.

Vector ac = a + c, vector ab = a + b, c = 1- ) a + b

Vector ac = a + 1 - ) a + b = a + b = a + b = a + b).

Vector ac= vector ab

Vector ac= vector ab

a, b, and c are collinear, and the end points of vectors a, b, and c are in a straight line.

Let there be a point on bc in obc, a (vector ob), b (vector oc), c = (vector oa), ab=kca, ab=ob-oa, ca=oa-oc into ab=kca, c-a=k(c-b), simplified to (k+1)c=a+kb, then =(1 k+1), k k+1), so + 1, THIS IS THE ORIGINAL METHOD. Your question is simple, bring =1- , into the above.

c= a+ b, the common factor is used to get the conclusion.

Then use the definition.

Let x1>x2, so y(x1)-y(x2)=x1 a 2-x2 a 2 be 1>x1>x2>0

Then the function decreases, so got.

y(x1)-y(x2)<0 decreasing.

The same goes for x1>x2>1.

y(x1)-y(x2)>0 increments.

Your problem is not justified, ah, hunger.

Sleep Peel Hill 8

The main thing is to look at x a 2. The old Xun in the back can not take the loose tube.

You see. It is an even function. Then a 2 should be even. a is again an integer.

So a is an even number. But in (0,+ is a subtraction function. So A can only be zero.

Think about it if a is an even number. So in (0,+ is not a subtraction function, you can take a few elimination numbers to see.

Solution: Connect AI, from the triangle rule of vector addition, AI=AB+BIFrom the question design, it can be seen that the point Q is the midpoint of the line segment AC, the point R is a third point of the line segment AB, the point A is AM BQ, and the extension line of the intersection CR is at the point M

From the knowledge of plane geometry, it can be seen that iq:am=1:2,am:

bi=1:2.∴iq:

bi=1:4.===>bi:

bq=4:5.Vector bi=(4 5)bq

Combined with the previous conclusions, it can be seen that ai=ab+bi=ab+(4 5)bq∴λ=4/5.

The title is problematic:

x 0, y 0, refers to the withered pin a 2 (xy) only travel (x+y) 0

The original formula can be reduced to (a-1)x-2 (xy)+ay 0, and both sides are divided by y, which yields: (a-1)(x y)-2 (x y)+a 0

Let (x y)=t, the inequality is converted to: (a-1)t -2t+a 0 is constant.

a-1>0 is required, the opening is facing up, and the discriminant formula =2 -4a(a-1) 0

i.e. a(a-1) 1, solution a (1 + 5) 2

If 2 (xy) is changed to 2 (2xy), then 2 a(a-1), the defeat is a 2

With your solution x=y, [x+2 (xy)] 2 (xy)=(x+2x) 2x=3 2, not 2+1 2

You didn't make it clear, for everything positive a, did you? This trillion-feast thing is nothing more than a few formulas to find the limit according to the inequality. (It is estimated that only you high school students still remember the clan Liang Yin) Zacha.

bx+2 xy+y x+y a, i.e. ( x+ y) x+y a, so (x* y) 2 x* y is a" = 2

I was wrong, and the method above was purely correct. First disadvantages).

x,y are both positive,x>0,y>0,(x+2 xy) (x+y) a, divide the left up and down by y at the same time, you have to change the year (x y+ hu su x y) (x y+1) a, and then use the commutation method to replace x y with t, (t+ t) (t+1) a, t>0, and then the method of finding the maximum value according to the year divination function can be.

tan2x=2tanx/(1-tan^2x)

Then all of them are transformed into tanx, and tanx is seen as a whole, 1 "filial piety Duan Chong tanx on his monotony.

Finding the pole of skillful annihilation is the burning cavity is its biggest straight point.

Let y=x

then f(2x)=f(x)-f(y)=0

That is, this is a constant function, and the value of the function is 0

That is, f(x)=0, which is the x-axis.

So it's both an odd and an even function.

Multiply the left side of the formula by sin 17, then the chain rent Qing formula becomes 1 shed grip 2*sin2 17*cos2 17*cos4 17*cos8 17, and so on, and finally becomes 1 16*sin(16 17), and sin(16 17)=sin 17, the formula begins to multiply by sin type royal 17, so the result 1 16*sin(16 17) is punished with sin 17, that is, the final answer is 1 16.