4 math problems in the second year of junior high school, 4 math problems in the second year of juni

Set a random number according to the conditions, and the eligible ones are (eg: a=5, b=2, c=7, d=0).

2.(x+5)÷2-1>(ax+2)÷2

x+5-2>ax+2

1-a)x>-1

x>-1/1-a

i.e. -1 1-a=

1=Solution: Add the first two formulas to get y=40-2x

Substituting the first equation is z=x-10

Since x, y, and z are all non-negative numbers, 40-2x 0 gets x 20x-10 0 gets x 10

So 10 x 20

m=5x+4y+2z

3( x+y+z)+(3x+y-z)-x

90+50-x

140-x, so 120 m 130

x=mx+1

x=-1/(m+1)<0

So m>-1

Root formula: x=(-b (b 2-4ac)) 2a solution: 1 9x^2-12x+4=9x

9x^2-21x+4=0

From the root finding formula, we can get: x1=(21+ 297) 18;x2=(21-√297)/18

2. (1/8)x^2+6x-16=0

From the root finding formula, we can get: x1=(-24+8 11); x2=(-24-8√11)

16x^2-48x-13=0

4x+1)(4x-13)=0

x1=-1/4;x2=13/4

4.From the root finding formula, we can get: x=-1 (1-m 2).

1 simplification to xy x-y -2 3

What is 2 y?

3. There is an objection. 4. There is an objection.

You can just bring it in !!

You can change the question again, no, there are some ambiguities in the question you wrote.

In the first question, a 3 + 3a = 1 can be obtained from the root that a is the third power of x + 3x-1=0

If a<0, a 3+3a<0.

If a>1, then there is a 3+3a>1

Therefore, we get 0 second question, f(x)=3x+7, so f(f(x))=3(3x+7)+7=9x+28

f(f(a))=100 so 9a+28=100 so a=8

Question 3, (1), f(x)=(m-2)x+2m-3=(x+2)m-2x-3

Let x+2=0 then there is f(x)=1 so the point (-2,1) must be passed

Fourth, because the straight line y=kx(k>0) passes the origin, and the hyperbola is also rotationally symmetrical with respect to the origin, then there is.

x1=-x2，y1=-y2

If the two points a and b are on the curve, then there is y2=4 x2 and y1=4 x1, which are substituted into the equation.

2x1y2-7x2y1=8（x1/x2）-28（x2/x1）=-8+28=20

(1) First of all, determine the approximate range of a according to the equation, a will definitely not be less than zero, because when a = 0, x 3+3x-1 is already less than 0, so a > 0. Then a must not be greater than 1, because when a=1, x 3 + 3x -1 is greater than 1. In this way, the key to determining 0 is y=ax+1-a to y-1 = a(x-1). This straight line is constant past (1,1) point, and the slope is a. Then according to the range of a above, draw a diagram.

The straight line does not cross the fourth quadrant.

2) Let's calculate f(a) :3f(a)+7=100 and find f(a) =31. Then find a : 3a+7=31. a=8

3)y = m-2)x+2m-3 is converted to y = x+2)m -2x -3 (-2,1), which means that no matter how much m is, when x=-2, m will be eliminated. x=1，y=4 ；x=2,y=5 or x=2,y=4;x=1，y=5。Try two cases, only the first case has a solution, m=3

4) Y=kx and y=4 x are coupled to give the binary linear equation kx 2 - 4 = 0. y2=kx2；y1=kx1, the substitution problem yields -5kx1x2 according to the root and coefficient relation x1x2 = 4 k substitution of the original formula = 20.

1.Knowing that a is a real root of the cubic of the equation x + 3x-1 = 0, then the straight line y=ax+1-a does not pass through the (4) quadrant.

ps I won't determine the size of 1-a <01-a<1

2.The function f(x)=3x+7, if f(f(a))=100, then a=(8).

3.The known function f(x) = (m-2)x+2m-3

1) Verify that no matter what real number m takes, these function images are constant over a certain fixed point.

f(x)=mx-2x+2m-3=(x+2)m-2x-3

When x=-2, no matter what value m takes, f(x)=1, so the fixed point (-2,1) is passed

2) When x changes in the range 1 "x" 2 and y changes in 4 "y " 5, find the value of the real number m.

f(1)=3m-5 f(2)=4m-7

f(2)>f(1), 4m-7>3m-5m>2 f(x) is the increasing function.

So 3m-5>=4, 4m-7<=5 m=3

f(2) 14 m no solution.

So m=34If the straight line y=kx(k>0) and the hyperbolic y=4/x intersect with a(x1,y1),b(x2,y2), then 2x1y2-7x2y1=(20).

x^3+3x-1=0;

x(x^2+3)=1

x^2+3=1/x>0；So x>0

x>0 on the left is the increment function;

There is a variable x>0 is a subtractive function;

When x=1, x2+3=4; 1/x=1;

So a must be less than 1;

i.e. 1>a>0; 1-a>0;does not pass through the fourth quadrant;

a.In the case of an acute triangle, you can use (when the right-angled side is half the hypotenuse, the corresponding angle of the side is 30 degrees, if you learn trigonometry, it is obvious that you can get it) to find the apex angle of 30 degrees, the base angle is 75°, and the base angle is 15 in the case of an obtuse triangle

Variance = 10 - ( 2 2) 2 = 19 2

ac) 2-(bc) 2=a 4-b 4(a 2-b 2)c 2=(a +b 2)(a 2-b 2)(a 2-b 2)(a 2+b 2-c 2)=0 is an isosceles triangle when a = b.

When a 2+b 2=c 2, it is a right triangle, and when a = b, and a 2+b 2=c 2, it is an isosceles right triangle.

Acute or obtuse triangles.

Variance = 10 - ( 2 2) 2 = 19 2

ac) 2-(bc) 2=a 4-b 4(a 2-b 2)c 2=(a +b 2)(a 2-b 2)(a 2-b 2)(a 2+b 2-c 2)=0 is an isosceles triangle when a = b.

When a 2+b 2=c 2, it is a right triangle, and when a = b, and a 2+b 2=c 2, it is an isosceles right triangle.

(1)√（5）²-2-√2|-3 times -27=5-(2-2)-(3).

2) 9 16 + 5 1 100==

3) A and b are opposite numbers to each other, then a+b = 0

c and d are the reciprocal of each other, then cd=1

x²+（a+b+cd）x+√(a+b)+³cd=x²+x+1

4) m is the integer part of 5, then m=2

n is the decimal part of 5, then n = 5-2

m-n=2-(√5-2)=4-√5