Knowing M N4, M2 N2 28, find the value of M N3 5

n+4)^2+n^2=28

2n^2+8n+16=28

n^2+4n+8=14

n+2)^2+4=14

n=(10^-2)-2

m=(10^-2)+2

m+n=2(10^-2)

m+n)^3=8*10^-6

x^2+(x^2-2x+1)=0

x^2+(x-1)^2=0

Because x 2>=0, and (x-1) 2>=0

x 2 and (x-1) 2 cannot be 0 at the same time

So there is no answer.

m-n=4

m^2+n^2=28

2mn=28-16=12

m+n)^2=28+12=40

m+n)^3=80*10^(-1)

Because ((-2) 2-4*2*1)<0

So there is no intersection with 0.

m-n=4m-n)^2=16

m^2+n^2=28

2mn=28-16=12

m-n)^2=16

m+n)^2=28+12=40

m+n) 3=80*10 under the root number

Knowing that 2x 2-2x+1=0, find the value of x.

The solution equation x does not exist.

Equation. The quadratic equation discriminant equation δ=b 2-4ac<0

So there is no real solution.

m 2 = n + 2 n 2 = m + 2, subtract the two formulas to get m 2-n 2 = n-m, because m ≠ n, so both sides are divided by m - n at the same time, to get m + n = -1, that is, the family hall n = -m - 1, bring in m 2 = n + 2, get.

m 2 = -m + 1 i.e. m 2 + m - 1 = 0, the same way n 2 + n - 1 = 0 so m, n are two different real roots of the equation x 2 + x-1 = 0, which are obtained by the relationship between the root and the sign of the hidden number such as: m + n = -1, mn = -1, so.

m^3+2mn+n^3=(m+n)(m^2-mn+n^2)+2mn=-[m+n)^2-3mn]-2=-[1+3]-2=-4

Known: m-2n = -2, 3-2m + 4n value collapse beam.

3-2(m-2n)

Friends who ask questions on the mobile phone can comment on the upper right corner of the client [satisfactory round imitation] orange shirt.

I'm glad to answer the questions for you, and I wish you progress in your studies! If you don't understand, you can ask!

If you need help with other topics, you can turn to me. Thank you!!

Whether it is m≠0 is m≠n

It is obtained by subtracting the two formulas of the source m 2 = n + 2 and n 2 = m-m, that is, (m-n) (m+n+1) = 0, and m≠n, so.

Take m-n≠0, so, m+n+1=0, and the solution is m+n=-1, so m3-2mn+n 3=m 3-mn+n 3-mn=m(m 2-n)+n(n 2-m)=2m+2n=2(m+n)=2 (-1)=-2

Subtracted from the two formulas m2=n+2 and n2=m+2, dum 2-n 2=n-m is (m-n)(m+n+1)=0, and m≠n, so m+≠n+1=0, and m+n+1=0, and the solution dao obtains m+n=-1, therefore.

Generic m 3-2mn+n 3=m 3-mn+n 3-mn=m(m 2-n)+n(n 2-m)=2m+2n=2(m+n)=2 (-1)=-2

From: 4 m*2 n-2=32:

2^2m×2^（n-2）=2^5

So: 2m+n-2=5

And because: m-2n=1, so: m=2n+1

So: 2(2n+1)+n-2=5

4n+2+n-2=5

n=1m=2×1+1=3

I'm glad to answer your questions and wish you progress in your studies!

Because |m|=2, |n|=3

So m=2 or m=-2, n=3 or n=-3

When m=2, n=3, m+n=2+3=5

When m=2, n=-3, m+n=2-3=-1, m=-2, n=3, m+n=-2+3=1, m=n=-2, n=-3=-5, m=n=-2-3=-5

m|=2, then m = 2

n|=3, then n = 3

So, m+n has four results, which are:

m-2n=-2, multiply both sides of the equation by -2 to get -2m+4n=4, and get -3-2m+4n=1