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Mathematical expansion questions with answers
(1)x is a rational number, and find y = |x- 100/22 1|+

The minimum value of |x+95/22 1|.

1, when x≥ 100/22 1

y = | x- 100/22 1 |+| x+95/22 1 |

= x- 100/22 1+95/22 1

= 2x - 5/22 1

The minimum value is obtained when x = 100/22 1.

ymin

= 2× 100/22 1 - 5/22 1 = 15/ 17

2. when-95/221≤ x <100/221

y = | x- 100/22 1 |+| x+95/22 1 |

=-x+ 100/22 1+x+95/22 1

= 195/22 1

= 15/ 17

3. when x

y = | x- 100/22 1 |+| x+95/22 1 |

= 100/22 1-x-x-95/22 1

= -2x + 5/22 1

The minimum value is obtained when x = -95/22 1.

ymin = 2×95/22 1

+ 5/22 1 = 285/ 17

So y = |x-

The minimum value of100/221|+x+95/221| is

ymin = 15/ 17

(2) 1/ 1+2+3+4+……+99+ 100 = 1/(n(n+ 1)/2)= 2( 1/n- 1/(n+ 1))

1+( 1/ 1+2)+( 1/ 1+2+3)+( 1/ 1+2+3+4+……+99+ 100)

=2( 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+..+ 1/ 100- 1/ 10 1)

=2( 1- 1/ 10 1)

=200/ 10 1

(3)

The distance from point x to a is |x-a|, and the distance from point x to b is |x-b|.

So: |x-a|=2|x-b|

① when x≥b≥a, x-a=2(x-b)=2x-2b.

So: x=2b-a

② when b≥x≥a, x-a =-2 (x-b) =-2x+b.

Therefore, x=(a+b)/3.

③ when b≥a≥x, there are: -(x-a)=-2(x-b)

Namely: -x+a=-2x+2b

So, x=2b-a

(4) The first maximum cube side length can only be 10 cm.

The second maximum cube side length can only be 18- 10 = 8cm.

The third maximum cube side length can only be 14-8 = 6cm.

(5) Let radius of the earth be R, the circumference is 2π r ... The length after increasing 1 m is (2π r+ 1) m,

The radius at this time is: (2π r+1) ÷ (2π) = r+1/2π.

Added gap: r+1/2π-r =1/2π m.

The radius of the globe is R. The circumference is 2π r. After increasing 1 m, the length is (2π r+1) m.

The radius at this time is: (2π r+1) ÷ (2π) = r+1/2π.

Added gap: r+1/2π-r =1/2π m.

According to the calculation, the wire will leave the surface of the earth and the globe, resulting in some gaps, and the gaps are the same. are all

(6) Let the length of this rectangular container be a..

10×3×a÷a÷4,

=30÷4,

=7.5 cm,

A: The height of water in the picture on the right is 7.5 cm.

(7) Assume that you can carry X kilograms of luggage for free, and the overweight part is Y yuan per kilogram.

( 150-3x)* y = 67+33+20 = 120

( 150-2x)* y = 88+92 = 180

xy=60

150y-3xy= 120

150y=300

y=2

The overweight part is 2 yuan per kilogram.

(8)3×3-6=3 (points);

Answer: The fourth time is 3 points more than the third time. Is that acceptable?