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Given the linear function y = (M-3) x + 2-N, when the value of Mn is, the intersection of function image and Y axis is above X axis
According to the meaning of the title
m-3≠0
2-n>0
∴m≠3,n
If f (x) = x ^ 2 + 3x + 1, then f (x + 1) is equal to
f(x+1)
=(x+1)²+3(x+1)+1
=x²+5x+5
Given that the function f (x) is equal to 3x plus 2, the value of F (2) is
Given that f (x) = 3x + 2, substitute x = 2 into f (2) = 3 * 2 + 2 = 8
1. Factoring in real numbers: x ^ 4-4=______
2. If the polynomial x ^ 2-2 (m-1) x + 64 is a complete square, then the value of M is_____
3. Carefully observe the equation (x + P) (x + Q) = (P + Q) x + PQ, if one factor of polynomial x ^ 2-5x + 6 is (x-3), then the other is_____
4. Factoring the polynomial x ^ 2-x + Y-Y ^ 2, the result is_____
5. Factorization 3AB (7c-28d) + 5ad (c-4d) - 2Ac (12d-3c)
6. Using factorization to solve the equation: 1 / 2x ^ 2 + + 1 / 2 = 0
1. Factoring in real number: x ^ 4-4 = (x ^ 2 + 2) (x + √ 2) (x - √ 2)
2. If the polynomial x ^ 2-2 (m-1) x + 64 is a complete square, then the value of M is (9 or - 7)
3. Carefully observe the equation (x + P) (x + Q) = x ^ 2 + (P + Q) x + PQ, if one factor of polynomial x ^ 2-5x + 6 is (x-3), then the other is (X-2)
4. Factoring the polynomial x ^ 2-x + Y-Y ^ 2, the result is (x + Y-1) (X-Y)
5. Factorization 3AB (7c-28d) + 5ad (c-4d) - 2Ac (12d-3c)
3ab(7c-28d)+5ad(c-4d)-2ac(12d-3c)
=21ab(c-4d)+5ad(c-4d)+6ac(c-4d)
=(c-4d)(21ab+5ad+6ac)
=a(c-4d)(21b+5d+6c)
6. Using factorization to solve the equation: 1 / 2x ^ 2 + + 1 / 2 = 0
Why are there two "+" numbers?
Exercises of group decomposition method in grade two of junior high school
1. Try to compare the values of polynomials (cubic power of M + product of square of M and N) and (product of square of M and N + cubic power of n)
2. Given that the positive integers m and N satisfy mn-m-n = 35, find the values of all m and N satisfying the condition
1. Subtraction = square of M (M + n) - square of n (M + n) = (M + n) (square of M - square of n)
=The square of (m-n) (M + n)
Therefore, when m > N, the previous formula is large
When m = n, it is equal,
When m
Given that a and B are integers and satisfy AB + A + B = 6, find the value of a + B
ab+a+b+1=7
(a+1)(b+1)=7
a+1=1,b+1=7
Or a + 1 = 7, B + 1 = 1
Or a + 1 = - 1, B + 1 = - 7
Or a + 1 = - 7, B + 1 = - 1
therefore
a=0,b=6
Or a = 6, B = 0
Or a = - 2, B = - 8
Or a = - 8, B = - 2
therefore
A + B = 6 or a + B = - 10
Four questions about factorization in grade eight
The third power of a (X-Y) - the second power of 3A B (n + 2b) then M =, n=
Factorization factor (x + 1) (x + 2) (x + 3) (x + 4) + 1=
Given X-Y = 1, find the quadratic power of X-Y + x-3y
When finding x + 3Y = 125 (x is not equal to y) by factorization, (the quadratic power of x = the quadratic power of 2xy-3y) divided by (X-Y)
I don't understand
2,
（x+1)(x+2)(x+3)(x+4)+1
={[x+1)(x+4)][(x+2)(x+3)]}+1
={[(x^2+5x)+4][(x^2+5x)+6]}+1
=(x^2+5x)^2+10(x^2+5x)+24+1
=(x ^ 2 + 5x + 5) ^ 2. (complete square formula)
3,
From X-Y = 1, so
x^2-y^2+x-3y
=(X-Y) (x + y) + x-3y (square difference formula)
=x+y+x-3y
=2(x-y)
=2.
4,
From x + 3Y = 125, so
(x^2+2xy-3y^2)/(x-y)
=(x + 3Y) (X-Y) / (X-Y) (cross multiplication factorization)
=x+3y
=125.
The following formulas are decomposed into factors: (1) - 4a3b2 + 8a2b2; & nbsp; & nbsp; & nbsp; (2) 9 (a + b) 2-4 (a-b) 2; & nbsp; & nbsp; (3) (x2 + Y2) 2-4x2y2
（1）-4a3b2+8a2b2，=-4a2b2（a-2）；（2）9（a+b）2-4（a-b）2，=[3（a+b）+2（a-b）][3（a+b）-2（a-b）]，=（5a+b）（a+5b）； （3）（x2+y2）2-4x2y2，=（x2+y2+2xy）（x2+y2-2xy），=（x+y）2（x-y）2．...
a²+b²+c²-ab-ac-bc=0
Syndrome a = b = C
Further proof of a + B + C = 0
A ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc = 02 (a ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc) = 0A ^ 2-2ab + B ^ 2 + B ^ 2-2bc + C ^ 2 + A ^ 2-2ac + C ^ 2 = 0 (a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0. Because the square of a number is non negative, A-B = 0, a = bb-c = 0, B = Ca-C = 0, a = C is a = C, then a = b = C
The square of X - 3x + 2
Do it by cross multiplication,
x^-3x+2
=(x-1)(x-2)
AI! It's so simple. Can you give me some difficult questions
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