Four math problems in the third year of junior high school owo 20

14 c = b-a

Original -->ax2 + bx + b - a = 0delta = b2 - 4ac = b2 - 4a(b-a) = b2 - 4ab + 4a2 = (b-2a)(b-2a).

x =( -b±(b-2a))/2a

x1 = -b-a \

x2 = -1

15 k != 0 delta >0

k!= 0 4(k+1)(k+1) -4k2 > 0k> and k!= 0

17 m = -5

2x2 - 5x - 3 = 0

Cross multiplication.

2x 11x -3

So (2x + 1) (x - 3) = 0x=3, x =

18 twice the beginning and tail in the middle, so |-a|= 2 * root number (2a - 3) a2 - 8a + 12 = 0

a-2)(a-6)=0

a = 2 a = 6

14 questions: -1 15 questions: 8k+4>0 17 questions: 18 questions: minus two root number three plus two two root number three plus two.

Both the third and fifth questions are b

1、∠c=x

As cd bc in d, cd=bcosx, ad=bsinxc 2-(a-bcosx) 2=bsin 2xa 2+b 2-2abcosx=c 2, it can be seen that at x=90°, c 2=a 2+b 2

x is an acute angle. c^2b^2+a^2

It's too tiring to do such a question in the third year of junior high school, sina

Make a right-angled triangle.

According to the equal area, the relationship between BC and AB, AC and its angle is introduced.

ab=bc*sin45/sin75,ac=bc*sin60/sin75

bc*(√2/4)*(2+√6)/sin75=2+√6bc=√2+√3

ab=ac=√3

Make a 15° 75° right-angled triangle so that the length of this side of the oblique potato = 2 is connected from the right-angled vertex to the midpoint of the inclined plane.

30°, 75°, 75° isosceles triangles are obtained, and in this triangle, the right-angled side length = (2-3) of 15° can be obtained by using the angle of 30 degrees

sin15°=√6/4-√2/4

1.When 1 a - 1 b - 1 (a+b) = 0, find the value of b a + a b.

b-a)/(ab)=1/(a+b)

ab=-(a-b)(a+b)=-(a^2-b^2)=b^2-a^2

Both sides are divided by ab:

1=b/a-a/b

The two sides are squared:

b/a)^2+(a/b)^2-2*b/a*a/b=1

b/a)^2+(a/b)^2=3

b/a+a/b)^2=(b/a)^2+(a/b)^2+2=3+2=5

So, b a + a b = root number 5 or - root number 5

2.Decomposition Factor:

x squared + x - 1

x^2+x+1/4)-5/4

x+1/2)^2-5/4

x+1 2 + root number 5 2) (x+1 2 - root number 5 2).

3 (x squared) -4 xy - y squared.

x^2-4xy+4y^2)-5y^2

x-2y)^2-5y^2

x-2y + root number 5y) (x-2y - root number 5y).

3.Knowing that the unary quadratic equation 2 (x squared) +3x - 5 = 0 is not solved, the unary quadratic equation is found with the reciprocal of the two roots of the equation as the root.

Vedic theorem:

x1+x2=-3/2

x1x2=-5/2

Let the two roots of the other equation be:

a=1/x1,b=1/x2

a+b=1/x1+1/x2=(x1+x2)/x1x2=(-3/2)/(-5/2)=3/5

ab=1/x1x2=-2/5

So the equation is: x 2-3 5x-2 5=0

Or: 5x 2-3x-2=0

4.In RTABC, cd = h is the height on the hypotenuse, bc = a ac = b

Verification: 1 (a squared) + 1 (b squared) = 1 (h squared).

Proven on the basis of equal area.

s(abc)=1/2*ac*bc=1/2ab.

and s(abc)=1 2*ab*cd=1 2*ch

So there is: 1 2ab = 1 2ch

i.e.: ab=ch

The square of the two sides gets: a 2b 2 = c 2h 2, c 2 = a 2 + b 2 substitution gets:

a^2b^2=(a^2+b^2)h^2

Divide by a 2b 2h 2

1 h 2 = (a 2 + b 2) (a 2b 2) = 1 b 2 + 1 a 2 certificate completed.

(1) From: 1 a - 1 b - 1 (a+b)= 0(1+b a)-(a b+1)-1=0

b/a-a/b=1

So: (b a + a b) 2=(b a-a b) 2+4=3b a + a b=root 5 or: b a + a b=-root 5(2) x square + x - 1=(x 2+x+1 4)-5 4=(x+1 4) 2-5 4

x+(1 2)+(root number5) 2)(x+1 2-(root number5) 2)(x squared) -4xy - y squared = (x 2-4xy+4y 2)-5y 2

x-2y)^2-5y^2

x-2y+(root number 5)y)(x-2y-(root number 5)y)(3) let x=1 y substitute 2(xsquared) +3x-5 = 0, get:

2(1/y)^2+3(1/y)-5=0

5y^2-3y-2=0

Replace the above y with x, then:

5x^2-3x-2=0

That's what you want. 4) The recorded area is s

2s=ac*bc=ab

2s)^2=a^2b^2

Again: 2s=ab*h

2s) 2=(ab) 2*h 2=(a 2+b 2)h 2 so: a 2b 2=(a 2+b 2)h 2 so: 1 (a squared) +1 (b squared) = 1 (h squared).

5. Sort out x=7y (3y-7)

If y is taken as an even number, the numerator is an even number and the denominator is an odd number; If y is taken as an odd number, the numerator is an odd number, and the denominator is an even number; So x can't be an integer. A should be selected.

180 (10*5) = RMB (advertising fee per unit area).

Yuan (advertising cost after expanding the area).

Advertising fees apply= 450 yuanA moment forever 523 will answer for you, I wish you learning and progress If you agree with my answer, please click the [Accept as satisfactory answer] button in time The mobile phone questioner can comment "satisfied" on the client

Isn't it three times the original advertising cost?

You can set a function so that x is the value to be evaluated, then 3000*(1+x)*(1+x)=3630 Solving x is the answer.