If x m x 2n 3 x m n and 4x 2 are the same term, and 2m 5n 6, find the value of 4m 2 2n 2

The x m-n in your question should be x (m-n), right?!

x m x 2n) 3 x m-n, where (x m x 2n) = x (m-2n), so (x m x 2n) 3=x (3m-6n);

x^m÷x^2n)^3÷x^(m-n)=x^(2m-5n);

From the problem (x m x 2n) 3 x m-n and 4x 2 are the same terms, we can see that the power of x is the same, therefore.

2m-5n=2 ——1）

and 2m+5n=6 – 2).

Combine (1) (2) to get m=2 , n = 2 5 so 4m 2-2n 2 = 4 * 4 - 2 * 8 25 = 384 25

x^m÷x^2n)^3÷x^(m-n)

x^(m-2n)]^3÷x^(m-n)

x^[3(m-2n)-(m-n)]

x^(2m-5n)

x (2m-5n) and 4x 2 are the same kind of terms.

x of the same number of times.

2m-5n=2

2m+5n=6

It should be 4m 2-25n 2 in the end, right?

4m 2-25n 2=(2m+5n)(2m-5n)=2*6=12I hope mine is helpful to you, o( o!

Solution: (xm x2n) 3 x m-n=(xm-2n) 3 x m-n=x 3m-6n x m-n=x 2m-5n, because it is the same term as 4x2, so 2m-5n=2, and 2m+5n=6, + get.

4m=8m=2 substitution to get n=2 5

If yes, find the value of 4m2-2n2.

4m^2-2n^2=2(2m^2-n^2)=2(8-4/25)=392/25

If so, find the value of 4m2-25n2.

Original formula = (2m+5n)(2m-5n)=2 6=12 I don't understand, please ask, happy o( o

First of all, we must figure out what is the same kind of term, the same kind of term refers to variables, two terms of the same degree, (x m x 2n) 3 x m-n can be reduced to x 3 (m-2n) x (m-n), here you should be missing a parenthesis, and then, simplify to x [3(m-2n)-(m-n)], and 4x 2 is the same kind of term, that is, 3 (m-2n)-(m-n)=2 that is, 2m-5n=2, and then according to the known conditions 2m+5n=6, m=2, n=2 5, can be solved.

xmx2n

XM-N (Rotten Royal XM-2N.)

xm-nx3m-6n

xm-nx2m-5n

Because it is with 4x2

is the same term, so 2m-5n=2, and 2m+5n=7, so 4m225n2(2m)2

Hitch old 5n) 2

2m+5n) (2m-5n), 7 2, Zhili Sheng.

The same kind of term is the knowledge point of the whole formula plus or minus the early fiber, and the same kind of trembling search item is two or more formulas with the same form and number of times of the whole calendar.

Let 2m-5n = 2

2m+5n=7

4m=9m=9/4

n=1/24m²-25n²=4×(81/16)-25×(1/4)

Because x 2m-5n and Qi Hao 4 2 are the same disturbance side space class term slow blindness, so 2m-5n=2, and 2m-ten5n=7, so 4m2-25n=(2m-ten5n)(2m-5n)=7x2=14

Answer: x m x 2n) 3 x (m-n)[x (m-2n)] 3 scramble x (m-n) x 3(m-2n) Sakura late x (m-n).

x^(3m-6n-m+n)

x^(2m-5n)

It is the same as 4x 2.

So 2m-5n=2

2m+5n=6

Solve the equation Spine Lie group.

m=2n=2 5

4m^2-25n^2

12,3,

x^m÷x^2n)^3÷x^(m-n)=x^3(m-2n)÷x^(m-n)=x^(3m-6n-m+n)=x^(2m-5n)

and -1 4x 2 for the same term.

Then 2m-5n=2 2m=5n+2

4m^2-2n^2=(5n+2)^2-2n^2=23n^2-20n+4=23(n-10/23)^2-8/23

It is impossible to ask for the final result, and there is a minimum value of the crack such as -8 23,1,

Kai Jan segment xmx2n

x2m-n（xm-2n

x2m-nx3m-6n

x2m-nxm-5n

Because it is related to 2x3

It is the same term, so m-5n=3, m+5n=13, m=8, and n=1, so m2 is changed

25n=82

If 5x 2 y m and 4x n+m+1 y 2n-2 are the same sock banquet hidden class hall auspicious hold, then.

n+m+1=2

m=2n-2

n=1m=0

From the definition of similar terms, we can see that m=2, 2n=6, so the hidden land n=3, then mn

Therefore, the answer is 8 stoves.

Let the polynomial be y=x +4x+6=(x+2) +2.

When x=2m+n+2, y=(2m+n+4) +2, and when x=m+2n, y=(m+2n+2) +2.

And because x=2m+n+2 and x=m+2n, the value of the polynomial x +4x+6 is equal, so (2m+n+4) +2=(m+2n+2) +2.

That is, 2m+n+4=m+2n+2 or 2m+n+4=-(m+2n+2), and because m-n+2≠0, 3m+3n+6=0, that is, m+n=-2.

So when x=3(m+n+1), the polynomial y=x +4x+6=(x+2) +2=(3m+3n+5) +2=[3x(-2)+5] +2=3.