1, it is known that n is a positive integer, 2n+ 1 and 3n+ 1 are completely squared, so:
n=40,
5n+3=5*40+3=203
Because 203=29*7 is not a prime number.
So there is no such number n; ? ##
2. Let m be an integer, and the equation MX 2+2 (m-5) x+m-4 = 0 about x has integer roots, so what is the value of m? (m=- 18)
= = & gtdelta:=[2(m-5)]^2-4m(m-4)= 100-24m
The solution of the original formula: x = [-2 (m-5) √ (100-24m)]/2m.
=- 1+[5√(25-6m)]/ m
=- 1+{5 √[5^2+(-6m)]}/m?
To make √ [5 2+(-6m)]} an integer,
= =>5 2+(-6m) must be a complete square number.
= => From Pythagoras number 5- 12- 13, we get
-6m= 12^2= 144
m =- 18;
= = & gt? x =- 1+{ 5 √[5^2+(-6*- 18)]}/(- 18)
=- 1+{5? √[5^2+ 12^]}/(- 18)
=- 1+(5 ? 13)/(- 18)
There is an integer root: =-1+(5+13)/(-18) =-2;
3. If for a natural number n not less than 8, when 3n+ 1 is a complete square number, n+ 1 can be expressed as the sum of k complete squares, then the minimum value of k is (? 1? )
a、 1? b、2C、3D、4
Solution:
When 3n+ 1 is a complete square number,? N+ 1 can be expressed as the sum of k complete squares,
A natural number n not less than 8, where n=8, with:
3*8+ 1=25 is a complete square number;
n+ 1 = 8+ 1 = 9;
9=3^2=2^2+2^+ 1^2;
So the smallest k =1; ?
4. If m 2 = n+2 and n 2 = m+2 (m is not equal to n), the value of m 3-2mn+n 3 is (? 0? )
a、 1B、0C、 1D、2
Solution: m 2 = n+2, n 2 = m+2, two expressions are subtracted: (m2-N2) =-(m-n) = > m+n =-1;
M 2 = n+2, n 2 = m+2, and the two expressions are added: (m2+N2) = (m+n)+4 = = > m 2+n 2 = 3;
Because: m+n =-1= = > (m+n)^2=(- 1)^2
= = & gt? m^2+n^2+2mn= 1
= = & gt? mn=[ 1-(m^2+n^2)]/2=( 1-3)/2=- 1;
m+n =- 1 = = & gt; (m+n)^3=(- 1)^3
= = & gt? m^3+n^3+3mn(m+n)=- 1
= = & gt? m^3+n^3= 1-3mn(m+n)= 1-3*(- 1)(- 1)=-2;
So: m3-2mn+n3 =-2-2 * (1) = 0; ? ##
5. Let N=23x+92y be a complete square number, and n does not exceed 2392, then all positive integer pairs (x, y)*** have _2 1 15_ pairs that satisfy the above conditions.
Solution: Because N=23x+92y,?
= = & gty=-x/4+N/92
Because n does not exceed 2392
So n/92 < = 2392/92 = 26;
In contrast, the possible range of N/92 (26, 25, 24, 23, 22…, 3, 2, 1) is: only when N/92=23, N/92=23.
= =>n = 2116 = 46 * 46, which is a complete square number.
= = & gty=-X/4+2 1 16
That is, find the positive integer solution (x, y) on the straight line y=-X/4+2 1 16.
= => The general solution of its positive integer:? (X=4K, Y=2 1 16-K), where (k is a natural number, K= 1, 2,3,, n).
To make Y=2 1 16-k a positive integer,
= = & gty = 2 1 16-k & gt; 0;
= = & gtK & lt2 1 16; K=2 1 15? ;
So * * * has 2 1 15 pairs of positive integers (x, y); ##
6. In the plane rectangular coordinate system xOy, we call the points whose abscissa is an integer and whose ordinate is a complete square number as "good points", and find the coordinates of all "good points" on the quadratic function Y = (x-90) 2-4907 image.
(Whether "4907" in the topic "Y = (X-90) 2-4907" is wrong, please check it carefully and modify it! ! ! )
7. It is known that the roots of the equation x 2-6x-4n 2-32n = 0 are all integers, so find the value of the integer n. ..
Solution: = => delta: = 6 2-4 (-4n 2-32n) = 36+4 (4n 2+32n).
Solution of the original formula: x = 6 √ [36+4 (4n 2+32n)]/2.
=3 √(4n^2+32n+9)
To make x an integer,
= = & gt4n 2+32n+9 must be a complete square number.
= = & gtGet: take 4n 2+32n+9 = (1,4,9,16,25,36,49,64, …, n 2).
4n^2+32n+9=9
= = & gtn = 0; ##
8. If D, E and F are points on BC, CA and AB of △ABC, respectively, BD: DC = 1, Ce: EA = 2, AF: FB = 3 and S△ABC=24, find the area of △DEF.
Solution:
(1) found S3.
△ABC, △AFC and △BFC have the same height with AB as the base and passing through point C, and are set as H.
So there is: AB*H=? s△ABC;
FB*H=? s△BFC; ?
Divided by two formulas: S△BFC=FB/AB*? s△ABC;
Because of AF? :FB = 3; ? = = & gtAB:FB = 4;
So: S△BFC=FB/AB*? s△ABC = 1/4 * 24 = 6;
At △BFC, d is the midpoint of BC, so:
S3 and S△DFC have the same area, = =>? S3=? s△BFC/2 = 6/2 = 3;
② Find S2, S 1?
△ABC, △ABE, and △BEC are based on AC and pass through point B, and their heights are the same, so they are set as Hb.
So there is: AC*Hb=? s△ABC; ? - (*)
AE*Hb=? S△ Abe; ? - (**)
EC*Hb=? s△BEC; ? - (***)
Divided by (*) and (* *): S△ABE=AE/AC*? s△ABC;
Divided by (*) and (* *): s △ BEC = CE/AC *? s△ABC;
Because CE:AE? =2; ? = = & gtAE:AC = 1/3;
= = & gtCE:AC = 2/3;
So: S△ABE=AE/AC*? s△ABC = 1/3 * 24 = 8;
S△BEC=CE/AC*? s△ABC = 2/3 * 24 = 16;
In △ABE, f is the (3: 1) point of AB, so: (Similarly, the height is equal, but the bottom is different).
The ratio of S2 to S△DFC area = the ratio of bottom edge =AF/FB=3: 1.
= = & gt? S2 and? The ratio of S△ABE = 3/4;
= = & gt? S2=? s△ABE * 3/4 = 8 * 3/4 = 6;
Similar S 1=? s△BEC * 1/2 = 16 * 1/2 = 8;
So s △ def = s △ ABC-s1-S2-S3 = 24-8-6-3 = 7; ##
9. Let a 2+1= 3a, b 2+1= 3b, a≠b, then the value of algebraic expression (1/a 2)+(1/b 2) is (. B=7? )
a、5B、7C、9D 1 1
Solution: A 2+ 1 = 3a, B 2+ 1 = 3b subtraction.
= = & gta^2-b^2=3(a-b)
= = & gt(a-b)(a+b)=3(a-b),? And a≠b,
= = & gta+b=3( 1)
A 2+ 1 = 3a,B 2+ 1 = 3b。
= = & gta^2+b^2+2=3(a+b)
= = & gta^2+b^2=3*3-2=7; (2)
Because (1)a+b=3.
= = & gt(a+b)^2=3^2=9
= = & gta^2+b^2+2ab=9;
= = & gt? 2ab=9-(? a^2+b^2)=9-7=2;
= = & gt? ab = 1; ; ?
So (1/a2)+(1/b2) = (a2+b2)/(? ab)^2=7/ 1=7; ? ##