Mathematics in the first year of high school, a simple process

5。If we know that the solution set of a= is, then a b=?

Solution: ax +bx+c=a(x+1)(x-2)=a(x-x-2)=ax -ax-2a>0, so a<0, and b=-a, c=-2a;

Substituting the expression of b yields b====

Hence a b =

b = with and only one element, then a b = ?

Solution: b==b= has and only one element;

Its discriminant formula =a -4=0, i.e., a = 2;That is, b = and a=, that is, for any x there is ax + (a-1)x+a-1 0, so there must be a<0;

And its discriminant formula =(a-1) -4a(a-1)=-3a +2a+1=-(3a -2a-1)=-(3a+1)(a-1) 0, i.e., there is (3a+1)(a-1) 0

Therefore a=== then a b= =.

Question 1: As can be seen from the solution set of a, a is definitely less than 0, and -1 and 2 are the two roots of a=0, bringing -1 and 2 into a=0 gives that b=-a, c=-2a. Then bring a, b, and c into b, and divide both sides by a at the same time (since a < 0, greater than 0 must become less than 0) to solve the solution set of b is -2, so anb=-1 The second problem: from b has and only one solution, it can be concluded that b=(x+a 2) 2-a 2 4+1=0 has only one solution, then a 2 4-1 = 0, so a = plus or minus 2. If a is true to a real number, it can be concluded that a must be less than 0

So a=-2i.e. anb = -2 (note that a and b are a set of a).Hope it helps.

In question 6, there is a set of b to know that a is equal to 2

Substituting A, verifying, can get the answer,

For (1+sina) cosa=-1 2, the numerator and denominator are multiplied by 1-sina at the same time.

1*1-sina*sina) cosa(1-sina)=-1 2 simplification yields cosa*cosa cosa(1-sina)=cosa (1-sina)=-1 2

So cosa (sina-1) = 1 2

By the equation of the doubling angle: cos2x=2cos x-1

Get: 1-2cos x = -2cos2x

So, y=-2cos2x

So, the period t=2 2=

The team of math enthusiasts will answer for you, if you don't understand, please ask and wish you progress in your studies!

<> is like silver, judging limbs, and trying to rush to the world.

a2 a4=a3 a3=1, so the deficit a3=1, so a1+a2=7-1=6, so a3 +a3 q=6, substitute a3=1, and an 0, so q=1 2, so a1=4, a2=2, a4=1 2, a5=1 no brother 4, pin cha roll.

So s7=4+2+1+1 2+1 4=

(1)lg5*(lg2+lg10)+(lg2)²=lg5*(lg2+1)+(lg2)²

lg5*lg2+lg5+lg2*lg2

lg2(lg5+lg2)+lg5

lg2*lg10+lg5

lg2*1+lg5

LG101(2) subscript will not be played. You'll just look at it, for example, log55 is preceded by 5 subscripts. The back is superscripted.

log5 5+log5 7-【log5 (7/3)²-log5 7+log5 9/5】

1+log5 7-【log5 49/9*1/7*9/5】=1+log5 7-log5 7

Let y=a*x 2+bx+c;

Past the origin, brother c=0;

The axis of the Xian family is x=1; So -b (2a)Zheng wither=1;

(1,2) so 2=a+b;

a=-2;b=4

Therefore y=-2a 2+4b

High school math is easy, and the process is not easy to type.