Middle School Math Problem Series 4 One is worth 50, a total of 2 questions, a total of 100 points

Q1: Known: y = |x-1| +x-3|1) Use the piecewise function to establish a Cartesian coordinate system and make an image of the function.

Note: It is necessary to clearly explain how to create the drawing.

When x-1=0, x=1

When x-3=0, x=3

1, when x<=1, y = |x-1| +x-3|=-(x-1)-(x-3)=-2x+42, when 13, y = |x-1| +x-3|=(x-1)+(x-3)=2x-4 is the same for plotting, and the images are drawn separately according to the intervals.

Q2: Use the vertical method to find the remainder of the polynomial 4x 8x 3x -7x divided by 2x+3.

Note: Indicates the power of 4; Indicates to the power of 3.

Again, explain it clearly.

4x^4+8x^3-3x^2-7x

4x^4+6x^3+2x^3+3x^2-6x^2-9x+2x-3+32x^3(2x+3)+x^2(2x+3)-3x(2x+3)+(2x+3)-3

2x 3+x 2-3x+1)(2x+3)-3 remainder is -3

Known: y = |x-1| +x-3|

1) Use the piecewise function to establish a Cartesian coordinate system and make an image of the function.

It is divided into 3 paragraphs. x>=3

y=x-1+x-3=2x-4

1<=x<3

y=x-1+3-x=2

x<1y=1-x+3-x=4-2x

Use the vertical method to find the remainder of the polynomial 4x 8x 3x -7x divided by 2x+3.

Note: Indicates the power of 4; Indicates to the power of 3.

4x^4+8x^3-3x²-7x=2x^3(2x+3)+x^2(2x+3)-3x(2x+3)+1(2x+3)-3

4x^4+8x^3-3x²-7x=(2x^3+x^2-3x+1)(2x+3)-3

The remainder is -3

1, "zero point" is x=1 and x=3.

When x<1, y=1-x+3-x=-2x+4, a ray with a slope of -2 is made from point (1,2) to the left.

When 13, y = x-1 + x-3 = 2x-4, make a ray with a slope of 2 from point (3,2) to the right.

And finally, an image of the " type.

2, original = 2x 2x+3)+2x 3x -7x2x 2x+3)+x *2x+3)-6x -7x2x 2x+3)+x *2x+3)-3x*(2x+3)+2x2x 2x+3)+x *2x+3)-3x*(2x+3)+(2x+3)-3

2x+3) (2x x -3x+1)-3 has a remainder of -3

It might help to see for yourself.

Go to the junior high school math website to find the test papers.

There are 25 big monks and 75 small monks.

One. Weigh the fifteen catties of oil tomatoes into the five-pound tube, and then put the five-pound barrel into the seven-pound one.

Two. Then put the twenty-pound bucket into the five-pound one, and then use the five-pound bucket to fill the seven-pound bucket.

Three. Put all the full seven catties into the twenty-pound barrel, and then put the remaining three catties of oil in the five-pound bucket into the seven-pound bucket.

Four. Trembling did put the twenty-pound cylinder into the empty five-pound barrel Qiaozhou, and then filled the seven-pound barrel with the five-pound barrel, and finally left a pound of oil in the five-pound barrel.

1. c

a + b + c = 0

a - b + c = 0

Subtract below above to get 2b = 0 --b = 0, and get a = -c, and substitute b into the equation to get ax 2 + c = 0

So x = plus or minus (-c a) = plus or minus 1

Choose C2 d

x = -a is a root, substituting it into the equation, get:

a)^2 - ab + a = 0

->a^2 - ab + a = 0

->a(a-b+1) = 0

Since a is not equal to 0, a-b+1 = 0, -a-b = -1 is constant, choose d

3. ba^2 + b^2 + c^2 + 50 = 6a + 8b + 10c

All items are moved to the left, resulting in (A 2 - 6A) +B 2 - 8B)+(C 2 - 10C) +50 = 0

Perfectly squared, we get (A 2 - 6A + 9) +B 2 - 8B + 16)+(C 2 - 10C + 25) = 0

So (a-3) 2 + b-4) 2 + c-5) 2 = 0

Because all of these squares are greater than or equal to 0, but their sum is 0, so they're all equal to 0

i.e. a-3 = b-4 = c-5 =0, so a=3, b=4, c=5

This triangle is a right-angled triangle, choose b, because according to the Pythagorean theorem: a 2 + b 2 = 3 2 + 4 2 = 9 + 16 = 25 = 5 2 = c 2

c. Substitution method.

d. Substituting a*a-a*b+a=0, since a is not equal to 0, approximate a, get a-b+1=0

b. The original formula is (a-3) 2+(b-4) 2+(c-5) 2=0, so a=3, b=4, c=5. Right-angled triangle.

(1) Because the equation a+b+c=0, a-b+c=0, both x=1 and x=-1 satisfy this equation, so there are at least two roots of this equation. And because this is a quadratic equation, there can only be two at most, so the root of the equation is x1=1, x2=-1

2) Because -a is the root of the equation, (-a) +b*(-a)+a=0i.e. a -ab+a=0

And because a≠0, the two ends of the above equation are reduced to a to get a-b+1=0So a-b is a fixed value: a-b=-1

Pick D. (3) By the title:

a +b +c +50-(6a+8b+10c)=(a -6a+9)+(b -8b+16)+(c -10c+25)=(a-3) +b-4) +c-5) =0The sum of the above three perfect squares is 0, so only a=3, b=4, c=5

At this point there is a +b =c, so abc is a right triangle.

1.As can be seen from the meaning of the question, bring in the options, and get the same equation as given, bring in x=1, get a+b+c=0, bring in x=-1, and get.

a-b+c=0.So the root of the equation is 1,-1. Choose C2Bring in x=-a, we get a*a-a*b+a=0, because a is not equal to 0, so we remove a a, we get a-b+1=0, a-b=-1

So a-b is constant, choose d

3.The original question is simplified to (A-3)*(A-3)+(B-4)*(B-4)+(C-5)*(C-5)=0

So a=3, b=4, c=5

So a*a+b*b=c*c

Therefore, the triangle is a right-angled triangle and B is selected

The answer is c, c, b

1) a+b+c=0 (1)

a-b+c=0 (2)

1)+(2) gets: a+c=0 solves: a=-c, b=0 so ax 2+bx+c=ax 2-a=0 i.e. x 2=1, so x= 1

2) Substituting (-a) into the equation yields: a 2-ab+a=0a=-b-1

Therefore, a+b=-1 is a constant.

3) Organize the original equation into (a-3) 2+(b-4) 2+(c-5) 2=0

So a=3, b=4, c=5

satisfies the Pythagorean theorem 3 2 + 4 2 = 5 2 The triangle is a right triangle.