Factorization of (x + y) ^ 2-3 (X-Y)

Factorization of (x + y) ^ 2-3 (X-Y)

(x+y)^2-3(x+y)
=(x+y)(x+y-3)
Find the value of - 12 + 36-108 +. + 4x3 to the power of 100
Let a = - 12 + 36-108 +. + 4x3 to the 100th power
=4×(-3)^1+4×(-3)^2+4×(-3)^3+…… +4×(-3)^100
=4×[(-3)^1+(-3)^2+(-3)^3+…… +(-3)^100]
Let B = - 3a
Then B = 4 × [(- 3) ^ 2 + (- 3) ^ 3 + (- 3) ^ 4 + +(-3)^101]
Let A-B = a + 3A = 4A
=4×[(-3)^1-(-3)^101]
That is 4A = 4 × [(- 3) ^ 1 - (- 3) ^ 101]
So, a = (- 3) ^ 1 - (- 3) ^ 101
=3^101-3
Mixed operation of fractions
2m2-4m/(2-m)(m-1)-(1+m)/1-m2
-1-a2/(a-1)-a
-1-a2/(a-1)-a
It's (- 1) - A2 / (A-1) - A
Don't read the title wrong
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)=-2m/(m-1)-1/(1-m)=(2m-1)/(1-m)(-1)-a^2)/(a-1)-a=(1-a-a^2-a^2+a)/(a-1)=-(2a^2-1)/(a-1)
Which is the bigger of 666 power of 3 and 333 power of 6
① ：6333=（2*3）333=3333×2333
② ：3666
① Then we get 3333 × 2333 / 3666 = 2333 / 3333 < 1
Therefore, ① ＜ ②, i.e. 6333 ＜ 3666
The 666th power of 3 is greater than the 333th power of 6
A few mixed operation problems of dichotomy in junior high school
Add and subtract, 15
It's urgent
A little more
2 a+1－a+3 a2－4a－5÷a2－9 a2－3a－10.
The original solution = [x + 2 x (X-2) - X-1 (X-2) 2] &; X 4-x
=[(x + 2) (X-2) x (X-2) 2-x (x-1) x (X-2) 2] &
=X2-4-x2 + x x (X-2) 2 &; x4-x (integral operation)
=X-4x (X-2) 2 &; x4-x
=X-4 x (X-2) 2 &; (- xx-4) (the sign rule of fraction)
=- 1 (X-2) 2
Calculate x + y x 2-xy + (x 2-y 2x) 2 and ﹥ 8226; (1 Y-X) 3
The original solution = x + y x (X-Y) + (x + y) 2 (X-Y) 2x2 &; 1 (Y-X) 3
=x+y x(x－y)－(x+y)2 x2(x－y)
=x2+xy－x2－2xy－y2 x2(x－y)
=－xy－y2 x2(x－y)=－xy+y2 x2(x－y).
x－y+4xy x－y）（x+y－4xyx+y）
Answer x2-y2
[1 (a+b)2－1(a－b)2]÷（1a+b－1a－b）
Answer 2A (a + b) (a-b);
x x－y• y2 x+y－x4y x4－y4÷x2 x2+y2
The answer - XY x + y
3x－2 x2－x－2+（1－1x+1）÷（1+1x－1）
The answer is X2 (x + 1) (X-2);
（2x x+1+2 x－1+4x x2－1）×（2x x+1+2 x－1－4x x2－1）.
Answer 4
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)
=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)
=-2m/(m-1)-1/(1-m)
=(2m-1)/(1-m)
(-1)-a^2)/(a-1)-a
=(1-a-a^2-a^2+a)/(a-1)
=-(2a^2-1)/(a-1)
On the inequality of X: 8 ^ x + 2 ^ X-2
Let 2 ^ x = t, obviously, there is t > 0  8 ^ x + 2 ^ X-2 = T ^ 3 + T-2 = (T ^ 3-1) + (t-1) = f (T) ∵ T ^ 3-1 = (t-1) (T ^ 2 + T + 1) ∵ f (t) = (T ^ 3-1) + (t-1) = (t-1) (T ^ 2 + T + 2) 8 ^ x + 2 ^ X-2 ＜ 0, that is, f (T) = (T ^ 3-1) + (t-1) = (t-1) (T ^ 2 + T + 2) ＜ 0 ∵ T ^ 2 + T + 2 ＞ 0
8 ^ x 2 ^ x is an increasing function, X
Mixed operation of fractions
1. Find the value of [x + 1 (X & sup2; - x) - x, X & sup2; - 2x + 1] / (X-2, X & sup2; - 2x), where x = - 1
2. First simplify, and then evaluate: 4x & sup2; - 4xy + Y & sup2; \ (2x + y) / (4x & sup2; - Y & sup2;), where x = 1, y = 1
[(x+1)/(x^2-x)-x/(x^2-2x+1)]/[(x-2)/(x^2-2x)]
=[(x+1)/x(x-1)-x/(x-1)^2]/[(x-2)/x(x-2)]
=[(x+1)(x-1)-x^2]/[x(x-1)^2]x
=-1/(x-1)^2
=-1/4
Original formula = [(2x-y) ^ 2 / (2x + y)] / (2x + y) (2x-y)
=(2x-y)/(2x+y)^2
=1/9
Binary first power mathematical problem?
2. This year, the number of students in school a and school B increased by 68 compared with last year, of which school a increased by 2% and school B increased by 5%. Now there are 1938 students in the two schools. How many students were there in school a and school B?
Let a be x and B be y 2% x + 5% y = 68 x (1 + 2%) + y (1 + 5%) = 1938 Should know how to calculate
The result of fractional blending must be
The result of mixed fraction operation must be the simplest fraction
Asking for help in Mathematics
1. The sum of two two digit numbers is 68. On the right side of the larger two digit number, write the smaller two digit number to get a four digit number. On the left side of the larger number, write the smaller two digit number to get another four digit number?
Let the large number be x and the decimal number be y
x+y=68
(100x+y)-(100y+x)=2178
The solution is: x = 45, y = 23
So these two double digits are 45 and 23