2X + 7 / 8 = 3 / 2 + X

2X + 7 / 8 = 3 / 2 + X

2X + 7 / 8 = 3 / 2 + X
2x-x=3/2-7/8
x=5/8

8 (the square of x-5x) = the square of 8x-7 and (3x-1) = the square of 4 (2x + 3)

8 (the square of x-5x) = the square of 8x-7, 8x & #178; - 40x = 8x-7x & #178; extract X; X (8x-40) = x (8-7x) x = 0, the equation holds, the square of 8x-40 = 8-7x15x = 48x = 16 / 5x = 0 or x = 16 / 5 (3x-1) = the square of 4 (2x + 3), then | 3x-1 | = | 4x + 6 | 3x-1 = 4x + 6 or 1-3x = 4x + 6x = - 7 or x = -

When k is a value, the solution of the equation 2x + k-1 = 1-x + k is 2

If so
(2x+k-1)/2=(1-x+k)/3
6x+3k-3=2-2x+2k
x=(5-k)/8=2 k=-11
If it is
(2x+k)/2-1=(1-x)/3+k
6x+3k-6=2-2x+6k
8x=8+3k
x=1+3k/8=2
x=8/3
If it is (2x + k) / 2-1 = (1-x + k) / 3
6x+3k-6=2-2x+2k
8x=8-k
x=1-k/8=2
k=-8

Suppose that two linear tables La and LB represent two sets a and B respectively. The new sets a = A and B are required
Data input:
\x05
\X05 by file input.txt The input data is given. The first behavior is the element of set a, and the second behavior is the element of set B
\X05 result output:
\x05
\X05 outputs the calculated elements in the new set a to the file output.txt .
\x05
Sample input file sample output file
\X05 input file example output file example
\x05 input.txt output.txt
\x052 6 3 9 8 6 2 6 3 9 8 6 5

#include
#include
#include
#include
using namespace std;
int main()
{
int a[]={1,5,8,12,5,-5,32};
int b[]={3,5,1,-3,10};
list< int > set1(a,a+sizeof(a)/sizeof(int));
list< int > set2(b,b+sizeof(b)/sizeof(int));
list< int > result;
set1.sort();
set2.sort();
//Intersection
set_ intersection(set1.begin(),set1.end(),set2.begin(),set2.end(),back_ inserter(result));
copy( result.begin (), result.end (),ostream_ iterator< int >(cout," "));
cout

Data structure algorithm 2.1 two linear tables La and LB, new set that, C language version of the complete program how to write ah

I wrote, you compile it, change it
Void Union (LinkList * L1, linklist * L2, linklist * & L3) / / intersection
{
LinkList *p=L1->next,*q=L2->next,*s,*c;
L3=(LinkList *)malloc(sizeof(LinkList));
L3->next=NULL;
c=L3;
while(p!=NULL&&q!=NULL)
{ if(p->data<q->data)
{s = (LinkList *) malloc (sizeof (LinkList)); / / copy node
s->data=p->data;
c->next=s;c=s;
p=p->next;
}
else if(p->data>q->data)
{ s=(LinkList *)malloc(sizeof(LinkList));
s->data=q->data;
c->next=s;c=s;
q=q->next;
}
else
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=p->data;
c->next=s;c=s;
p=p->next;
q=q->next;
}
}
while(q!=NULL)
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=q->data;
c->next=s;c=s;
q=q->next;
}
c->next=NULL;
while(p!=NULL)
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=p->data;
c->next=s;c=s;
p=p->next;
}
c->next=NULL;
}

C language programming problem, using two linear tables La and lb to represent two sets a and B respectively, now requires a set a = A and B

The trial position of la-abl-c L B L is shown in the known graph
Location: a --- 0 --- B -- C

As can be seen from the number axis, a-b

The positions of a, B and C on the number axis are known as shown in the figure below, which is simplified as | A-B | + | B + C | - | a-c |

It can be seen from the number axis that a ＜ B, a ＞ C, A-B ＜ 0, a-c ＞ 0, and B ＞ 0, C ＜ 0, B + C ＜ 0, so the original formula = - (a-b) - (B + C) - (A-C) = - A + b-b-c-a + C = - 2A

The corresponding points of rational numbers a, B and C on the number axis are shown in the figure. Try to simplify the algebraic formula 3ia Bi + Ia + bi IC AI + 2ib CI
---c-----------b------0---a-----→

3Ia-bI+Ia+bI-Ic-aI+2Ib-cI.
=3(a-b)-(a+b)+(c-a)+2(b-c)
=3a-3b-a-b+c-a+2b-2c
=a-2b-c

Given la-1l + LB + 2L = 0, find the power of a + B + (a + b) square +... + (a + b) to the power of 2001 + (a + b) to the power of 2012

a-1=0, a=1
b+2=0, b=-2
a+b=-1
A + B + (a + b) square +... + (a + b) to the power of 2001 + (a + b) to the power of 2012
=-1+1-1+1...-1+1
=0