Two math problems Help! Ask two math questions urgently!

1. Solution: According to the meaning of the question, it is obtained by Veda's theorem.

x1+x2=-1 3,x1*x2==-1 3, so 1 x +1 x =1

2. Solution: Since the point a is on the function y=4 x, then x *y =4 is the area of the rectangle is 4

And because the point a is on the straight line y=6-x, so x +y = 6 so the circumference of the rectangle is 2*6=12.

1. Find the values of x +x and x *x.

1 x +1 x =(1 x +1 x) (1 x *1 x)2, these two equations should be combined, there should be two solutions, and the coordinates of the first quadrant are the length and width of the rectangle.

Do the rest yourself!

x +1 x =(x1+x2) x1x2=1 xy=4;x+y=6

Area=x1*y1=4

Circumference = 2 * (x1 + y1) = 2 * 6 = 12

Categories: Education, Science, >> Learning Aid.

Problem description:1The real number x y z satisfies x+y+z-2(xy+yz+zx)+4xyz= at least one of the proofs is exactly equal to.

Hint Verification:(

2.Knowing a+b=1, a2+b2=2, find the value of a7+b7.

The number means to the power of the word.

Help me out, thanks!

Analysis: 1) Solution: Prove: Because x+y+z-2(xy+yz+zx)+4xyz=

Multiply both sides by 2 at the same time.

2x+2y+2z-4xy-4yz-4zx+8xyz=1

Shift: 2x-1+2y-4xy+2z-4zx-4yz+8xyz=0

Mention common factor: (2x-1)+2y(1-2x)+2z(1-2x)-4yz(1-2x)=0

Again: (2x-1)(1-2y)(1-2z)=0

The original question is proven. 2) a+b=1, a+b=2, so ab=-1 2

a3+b3)=(a+b)3-3ab(a+b)=1+3/2=5/2

a4+b4)=(a^+b^)^2a^b^=7/2

So (a7+b7)=(a3+b3)(a4+b4)-a3b3(a+b)=35 4+1 8=71 8

The fault of the first bit is in solving a=1 or -1 2

The second fault is that he is not a calculation at all, but a guess.

1.Based on 160 cm, the average height is.

160+(2+1-3-4+3+4+9-7+1-5+6-1) 12=cm.

2 a<0 b<0 |a|>|b|, then a< B A-B<0< P>

1.Using 160 as a benchmark, the height of each student is calculated as +2, +1, -3, -4, +3, +4....Then the average value is calculated, and then added to 160 to get the average height.

Question 1. For example, if you set 160 as the standard height. Then 12 students: 2, 1, -3, -4....

And then you can figure out the average of those 12 numbers, and add 160 and you're OK!

Question 2. a-b<0

n -n = n(square of n-1) =n(n+1)(n-1) One of the three natural numbers must be a multiple of 3, and at least one of the numbers is even, and the product is divisible by 6.

Similarly, n 5-n = n(n 4-1) = n (n 2 + 1) (n + 1) (n - 1), while 120 = 4*5*6,

1. Find the least common multiple of 7 and 8 and add 1Column: 7 8+1=57 (pcs).

2. No, you cannot. Because each person flips 4 times, the final result is an even number, and the three cups are reversed to an odd number, so it can't be done.

1.7 * 8 = 56, it should be 57.

2.No, because of the three cups, each person flips 4 times, odd and even problems.

1。7*8+1=57.

2。If you can't set the cup mouth up to + and the cup mouth down to -, then the lifetime arrangement is to first change the three numbers to the number It takes three times to make all the cups number 3 2n times 4k is not eligible.

The first question is 57, multiples of 7 and 8 are 56, and adding 1 is 57

Question 2 is no, because there are only 4 cups and 1 or 6 cups when you move 4 cups at a time to turn the rim down.

1. Divide some peaches evenly among 7 monkeys, there is exactly one left, and there is exactly one left among 8 monkeys, please ask if there are at least 56 + 1 = 57 peaches.