Mathematics problems in junior high school, please ask everyone!

AD is the bisector of the inner angle.

So. cad=∠bad

de//ac

So cad= ade bad= ade, so the triangle ade is an isosceles triangle ae=de

So. de/ac=be/ab=ae/ac

Proportional theorem. de/ac=(be+ae)/(ab+ac)=ab/(ab+ac)=3/5

Because it is parallel, so the angle eda = angle dac = angle ead, so ae = ed

Let ae=ed=x

Because of parallel, be ab=ed ac

be=ab-ae=3-x, so (3-x) 3=x 2

So x=6 5

So de ac=x 2=3 5

That's detailed enough!

The process is complicated, so I won't write about it.

2、（a-1）（a²+a+1）（a^6+a^3+1);This is done with the cubic variance formula.

a^3-1)(a^6+a^3+1)=a^9-1

Drawing the graph, it can be concluded that the total area of the parallelogram AEPF, BFP, PEC is equal.

Parallel proportional, so there is s bfp=x 2 4, s pec=(2-x) 2 4

s△abc=1

s parallelogram aepf = 1-x 2 4-(2-x) 2 4 = -x 2 2+x

The area of the triangle is half the area of the parallelogram.

s pef = s parallelogram aepf 2 = -x 2 4 + x 2 = -1 2 (x-1) 2 + 1 2

When x = 1, s pef max = 1 2

So when P is at the midpoint of BC, the S pef is maximum, and the maximum is 1 2

Solution (1) Do EH BC FG BC

ep‖ab△epc∽△abc

According to the ratio of the corresponding height of similar triangles, it is equal to the similarity ratio, which can be obtained.

eh/ad=pc/bc

Let bp=x and pc=2-x

Substitution is available. eh=1 - x/2

The same can be said. fg=x/2

ab‖ep ac‖fp

Quadrilateral AFPE is a parallelogram.

s pef = 1 2 quadrilateral afpe

s△abc-s△fpb-s△epc

1-1/2x*x/2-1/2(2-x)*(1 - x/2)=-1/2x²+x (0＜x≤2)

2) The quadratic function of the triangle area opens downwards and has a maximum value, which can be obtained according to the vertex coordinate formula.

When x=1, there is a maximum.

s pef max = 1 2

That is, when the point p is the midpoint of BC, the triangle area has a maximum value

abc, fbp, epc are similar, high ratio bottom: 1:2, s fbp = x * (x 2) divided by 2, = 4 x squared, s epc = (2-x) * (2 out of 2 (2-x)) divided by 2 = 4 squares (2-x), s pef = 2 / [1-4 x squared - 4 squares (2-x) square] = -1 4 * [x-1 squared] + 3 4

2, x = 1, s is the maximum, so at the midpoint of BC.

s△pef=(s△abc-s△bpf-s△pec)/2s△abc=1*2*1/2=1

s△bpf=x^2/4

s△pec=(2-x)^2/4

s pef=(1-x 2-(2-x) 4) 2 The apex of the parabola is x

x+y+xy=5 (1)(x 2)y+x(y 2)=6 (2)(2).

xy(x+y)=6 （3）

and (1) to solve the system of equations.

xy)^2+5xy+6=0

Think of xy as an unknown solution to a quadratic equation.

Get xy=2 or xy=3

When xy=2 (4).

x+y)=3 (5)

Solution (4) (5).

There is no real solution. When xy=3 (6).

x+y=2 (7)

The solution yields x=1, y=2

or x=2 , y=1

(x+y+xy)x=5x 1(x+y+xy)y=5y 2 condition: 1=x 2+xy+(x 2)y=5x

Condition 2 = xy + y 2 + x (y 2) = 5y, so condition 1 + condition 2 = (x + y) 2+ (x 2) y + x (y 2) = 5x + 5y

i.e. =(x+y) 2+6=5(x+y) then you can do the math yourself)

a*a*a*a + b*b*b*b >=2a*a*b*b if and only if a=b.

c*c*c*c + d*d*d*d >=2c*c*d*d if and only if c=d.

2*a*a*b*b+ 2*c*c*d*d >= 4*a*b*c*d if and only if a*b=c*d.

That is, a exp4 +b exp4 +c exp4 +d exp4 >=4abcd if and only if a=b=c=d, take the equal sign.

a 4 + b 4 + c 4 + d 4 > = 2a 2 * b 2 + 2c 2 * d 2 (if and only if a = b and c = d the equal sign holds).

4abcd (if and only if ab=cd the equal sign is true), because a, b, c, d are all positive rational numbers, so a=b, c=d, ab=cd can be deduced a=b=c=d, at this time the equal sign is true, that is.

But this seems to be the content of the first year of high school, a 2 + b 2 > = 2ab application).

A 4 + B 4 + C 4 + D 4> = 4ABCD (the equal sign holds if and only if A = B = C = D).

The easy way to do this should be this: Move item to get.

1 (x+3)-1 (x+4)=1 (x+1)-1 (x+2) The left and right sides are divided into the right and left sides

1 (x+3)(x+4)=1 (x+1)(x+2)j i.e.: (x+3)(x+4)=(x+1)(x+2) so there is 7x+12=3x+2

So x=-5 2

Multiply the common denominator on both sides.

x+1)(x+3)(x+4) +x+1)(x+2)(x+4) = (x+2)(x+3)(x+4) +x+1)(x+2)(x+3)

And then directly. 2 x^3+15 x^2+33 x+20 = 2 x^3+15 x^2+37 x+30

Can be solved. x = -5/2.

I don't know if it's right or not, I think I've completely forgotten...(Solution: 1 (x+2+x+3)=1 (x+1+x+4); 1/(2x+5)=1/(2x+5)=1/(2x+5); 0=1/(2x+5)-1/(2x+5);0=1/[(2x+5)-(2x+5)];0=1/(2x+5-2x-5);0=0.

Go to the denominator to get:

x+1)(x+3)(x+4) +x+1)(x+2)(x+4) = (x+2)(x+3)(x+4) +x+1)(x+2)(x+3)

Using the multiplicative distributive property and merging similar terms, we get:

2 x^3+15 x^2+33 x+20 = 2 x^3+15 x^2+37 x+30

Since there are 2 x 3+15 x 2 on both sides, it can be made to go, so that there is: -4x=10

Therefore x=-5 2

Move the item first, then pass the divide, and cross out the denominator to calculate.

Get rid of all the denominators, and then you're going to do it with the commutation method.

When x > 0, it is possible to convert the original to 1+1 x, where 0<1 x < 1, so, 1

Primitives are 1 x+11 x is an inverse function, when x is greater than 0, the value range is greater than 0, plus the previous 1, the value range of the original formula is greater than 1.

For x 2-3x-1 = 0, since x is not equal to 0, divide x by x on both sides at the same time to get x-(1 x)=3

So (x-(1 x)) 2=9

i.e. x 2 + (1 x 2) = 11

+2 on both sides of the equation gives (x 2+(1 x 2)+2=13, i.e. (x+(1 x)) 2=13

x+(1 x) = root number 13

Substituting x 2 + (1 x 2) = 11 and x + (1 x) = root number 13 into the above equation gives 12 times the root number 13

2.（x^2+2xy+y^2)(x^2-xy+y^2)^2=（x+y）（x+y）（x^2-xy+y^2）^2=(x^3+y^3)^2

1.The solution of x -3x-1 = 0 is not zero.

Divide both sides by x at the same time.

x-1/x=3

x³+1/x³=（x+1/x)（x²-1+1/x²)=（x+1/x)[（x+1/x)²-3]

(x+1 x) =(x-1 x) +4=13 x +1 x =10 13

2.（x²+2xy+y²)(x²-xy+y²)²=（x+y²)(x+y) (x²-xy+y²)=(x+y)(x²-xy+y²)(x+y) (x²-xy+y²)=（x³+y³）²