（2/5）X－（1－2/5）X/（1＋3/5）＝56

（2/5）X－（1－2/5）X/（1＋3/5）＝56

（2/5）X－（1－2/5）X/（1＋3/5）＝56
(2/5)*x-3/5*x/(8/5)=56
(2/5)*x-(3/5)*(5/8)*x=56
(2/5)*x-(3/8)*x=56
(16/40)*x-(15/40)*x=56
(1/40)*x=56
x=56*40=2240

It is known that the square + mx-n factorization factor of the binomial trinomial of X is (x + 4) (X-2), then the values of M and N are

The original formula is X & # 178; + mx-n = (x + 4) (X-2) (expansion)
=x²＋2x－8
Comparing the left and right coefficients, it is easy to get
m=2 n=8

56 + 4x = 84 x = how much

56+4X=84
4x=84-56=28
x=28/4=7
That is, x = 7

It is known that the bivariate cubic term of X is 2x & # 178; + MX + n. The result of factorization is to find the value of M and N by one fourth (2x-1) (4x + 1)~

1/4*(2x-1)(4x+1)
=1/4(8x²-2x-1)
=2x²-1/2*x-1/4
So m = - 1 / 2, n = - 1 / 4

Find x 4x square = 9, 16x square - 8 = 56, 9 [X-2] square = 25, 4 [x + 3] square + 4 = 125 in the following formulas

4X square = 9
2x=±3
x=±3/2
16x square-8 = 56
16x²=64
x²=4
x=±2
9 [X-2] square = 25
(x-2)²=25/9
x-2=±5/3
x=2±5/3
X = 11 / 3 or x = 1 / 3
4 [x + 3] square + 4 = 125
(x+3)²=121/4
x+3=±11/2
x=-3±11/2
X = - 17 / 2 or x = - 5 / 2

What is the value of a + B if the square + AX-1 of binary trinomial x can be decomposed into [X-2] [x + b]

x^2+ax-1
=(x-2)(x+b)
=x^2+(b-2)x-2b
a=b-2
-1=-2b
b=0.5,a=-1.5
a+b=-1

1/20+1/30+1/42+1/56
Simple calculation,

Solution
simple form
=(1/4-1/5)+(1/5-1/6)+(1/6-1/7)+(1/7-1/8)
=1/4+(1/5-1/5)+(1/6-1/6)+(1/7-1/7)-1/8
=1/4-18
=1/8

How to do the factorization of X ＾ (M + 2) - 2x ＾ (M + 1) - x ＾ m?

Finally, it's + X ＾ M
Original formula = x ＾ m * X & ＾ 178; - 2x ＾ m * x + X ＾ M
=x＾m(x²-2x+1)
=x＾m(x-1)²
If it's really - x ＾ m
Then the original formula = x ＾ m (X & # 178; - 2x-1)

One in 20 plus one in 30 plus 42, one in 56, only one in 72

One in 20 plus one in 30 plus 42, one in 56, only one in 72
=1 / 4 minus 1 / 9
=5 out of 36

If 2,3 are the two roots of the quadratic equation AX & # 178; + BX + C, then the factorization factor ax & # 178; + BX + C =?

2,3 are the two roots of the quadratic equation AX & # 178; + BX + C
x=2,x=3,
a(x-2)(x-3)=0
ax²+bx+c=0
ax²+bx+c=a(x-2)(x-3)