How to do these questions: 36a2-25b2-10b-1 question 2: B2 + 4a-1-4a2

How to do these questions: 36a2-25b2-10b-1 question 2: B2 + 4a-1-4a2

1、
Original formula = 36a & sup2; - (25B & sup2; + 10B + 1)
=36a²-(5b+1)²
=(6a+5b+1)(6a-5b-1)
2、
The original formula = B & sup2; - (4a & sup2; - 4A + 1)
=b²-(2a-1)²
=(b+2a-1)(b-2a+1)
（x+y）/（a-b）=（y+z）/（b-c）=（z+x）/（c-a）=k
x+y=k(a-b)
y+z=k(b-c)
z+x=k(c-a)
Add
2(x+y+z)=k(a-b+b-c+c-a)=0
x+y+z=0
The factorization of the second year of junior high school. We need to be more detailed about the class. Good must be added
1.4x^2-4x^2y-y^3
2.3x^3-18x^2+27x
3.(a-b)^2-12（a-b)c+36c^2
4.(x+y)^2-4(x+y-1)
5.(x+1)（x+2)+1/4.
1.4x ^ 2-4x ^ 2y-y ^ 3 should be 4xy ^ 2-4x ^ 2y-y ^ 3 = - Y (4x ^ 2-4xy + y ^ 2) = - Y (2x-y) ^ 22.3x ^ 3-18x ^ 2 + 27x = 3x (x ^ 2-6x + 9) = 3x (x-3) ^ 23. (a-b) ^ 2-12 (a-b) C + 36C ^ 2 = [(a-b) - 6C] ^ 2 = (a-b-6c) ^ 24. (x + y) ^ 2-4 (x + Y-1) = (x + y) ^ 2-4 (x + y) + 4 = [(x + y) - 2] ^ 2
1. The original formula = - Y (4x ^ 2-4xy + y ^ 2) = - Y (2x-y) ^ 2
2. The original formula = 3x (x ^ 2-6x + 9) = 3x (x-3) ^ 2
3. The original formula = (a-b) ^ 2-12 (a-b) C + (6c) ^ 2 = (a-b-6c) ^ 2
4. The original formula = (x + y) ^ 2-4 (x + y) + 2 ^ 2 = (x + Y-2) ^ 2
5. The original formula = x ^ 2 + 3x + 2 + 1 / 4 = x ^ 2 + 3x + 9 / 4 = (x + 3 / 2) ^ 2
(1) The original formula = - Y (4x ^ 2-4xy + y ^ 2) = - Y (2x-y) ^ 2
（2）=3x(x^2-6x+9)=3x(x-3)^2
(3) Let's take A-B as a whole and set it as a (for your understanding) original formula = a ^ 2-12ac + 36C ^ 2 = (a-6c) ^ 2
=(a-b-6c)^2
(4) Divide the - 1 of (x + Y-1) into the original formula = (x + y) ^ 2-4 (x + y) + 4. As in the above question, x + y is regarded as a whole, and the original formula = (x + Y -... Is expanded
(1) The original formula = - Y (4x ^ 2-4xy + y ^ 2) = - Y (2x-y) ^ 2
（2）=3x(x^2-6x+9)=3x(x-3)^2
(3) Let's take A-B as a whole and set it as a (for your understanding) original formula = a ^ 2-12ac + 36C ^ 2 = (a-6c) ^ 2
=(a-b-6c)^2
(4) Divide the - 1 of (x + Y-1) into the original formula = (x + y) ^ 2-4 (x + y) + 4. As in the above question, x + y is regarded as a whole, and the original formula = (x + Y-2) ^ 2
(5) Let x + 1.5 = a, the original formula = (a-0.5) (a + 0.5) + 0.25 = a ^ 2-0.25 + 0.25 = a ^ 2, substitute, the original formula = (x + 0.5) ^ 2
It's all brought in by practice
The complete square formula LZ used in it should be no problem
I've got a plan. LZ chooses me to put it away
Is there a problem with this question
2.3x^3-18x^2+27x
=3x(x^2-6x+9)
=3x(x-3)^2
3.(a-b)^2-12（a-b)c+36c^2
=(a-b-6c)^2
4.(x+y)^2-4(x+y-1)
=(x+y)^2-4(x+y)+4
=(x+y-2)^2
5.(x+1)（x+2)+1/4
=x^2+3x+9/4
=(x+3/2)^2
Using formula method to solve equation (1) 8x & # 178; - 7x + 3 = 0 (2) M & # 178; - 4m + 4 = 0
（1）8x²－7x＋3=0
△=b²-4ac= 7²－4*8*3 =-47
If the second m + 1 power of the second power B of 5A is the same as the second m + 3 power of the second power B of negative 2a, then M=
2m+1=m+3
M=2
（1）（-2x²y³)².(xy) · (xy)³
(2)(2a+3b)(2a-b)
(3)5x²（x+1）（x-1）
（4）（2x+y-1）²
（5）59.8×60.2
（6）198²
(7) (2a) & B 4
(8) (- 2 / 3 a seven B five) △ 3 / 2A & # 178; B five
(9) (6 / 5 A & #179; X four - 0.9ax & #179;) △ 3 / 5AX & #179;
（10）（7x²y³-8x³y²z）÷8x²y²
（11）12⒀÷（3⑽ · 4⑾）
(12) (factorization)
1.25x²-16y²
2.（a-b）（x-y）-（b-a）（x+y）
3.a²-4ab+4b²
4.4+12（x-y）+9（x-y）²
Hope there is a process, if you feel trouble can also write a short process or not
If I can't type it to the 13th power, I will use ⒀ and other symbols
(1)(-2x²y³)².(xy) • (xy)³ =4x^8 y^10
(2)(2a+3b)(2a-b)=4a²-2ab+6ab-3b²=4a²+4ab-3b²
(3)5x²(x+1)(x-1)=5x²(x²-1)=5x^4 -5x²
(4)(2x+y-1)² =(2x+y) ²-2(2x+y)+1= 4x²+4xy+y²-4x-2y+1
(5)59.8×60.2=(60-0.2)*(60+0.2)=60²-0.2²=3600-0.04= 3599.96
(6)198²=(200-2) ²=200²-2*2*200+2²=40000-800+4=39204
(7)(2a) ²•b^4=4a² b^4
(8) (- 2 / 3 A ^ 7b ^ 5) / / 3 / 2A & # 178; B ^ 5 = (- 2 * a ^ 7 B ^ 5 * 2A & # 178; B ^ 5) / (3 * 3) ---- a seven B five is not the denominator, right
= -4/9( a^9 b^10)
=(- 2 * 2A & # 178; B ^ 5) / (3 A ^ 7 B ^ 5 * 3) = - 4 / (9 A ^ 5) ---- a 7 b 5 is the denominator
(9)（6/5 a³ x^4-0.9ax³）÷3/5ax³=2a²x - 1.5
(10)（7x²y³ -8x³ y²z）÷8x²y²=(7/8) y - xz
(11)12^13÷(3^10•4^11)=(3*4)^13÷(3^10•4^11)=3³•4²=27•16=432
(12) (factorization)
1. 25X & # 178; - 16y & # 178; = (5x) & # 178; - (4Y) & # 178; = (5x-4y) (5x + 4Y) ---- square difference
2.(a-b)(x-y)-(b-a)(x+y)=(a-b)[(x-y)+(x+y)]=2x(a-b)
3.a²-4ab+4b²=a²- 2•a•2b+(2b) ²=(a-2b) ²
4. 4+12(x-y)+9(x-y)²=2² + 2*2*3(x-y) +9(x-y)²=[2+3(x-y)]²=(3x-3y+2) ²
I hope you will adopt it
(1)(-2x²y³)².(xy) • (xy)³ =4x^8 y^10
(2)(2a+3b)(2a-b)=4a²-2ab+6ab-3b²=4a²+4ab-3b²
(3)5x²(x+1)(x-1)=5x²(x²-1)=5x^4 -5x²
(4) ... unfold
(1)(-2x²y³)².(xy) • (xy)³ =4x^8 y^10
(2)(2a+3b)(2a-b)=4a²-2ab+6ab-3b²=4a²+4ab-3b²
(3)5x²(x+1)(x-1)=5x²(x²-1)=5x^4 -5x²
(4)(2x+y-1)² =(2x+y) ²-2(2x+y)+1= 4x²+4xy+y²-4x-2y+1
(5)59.8×60.2=(60-0.2)*(60+0.2)=60²-0.2²=3600-0.04= 3599.96
(6)198²=(200-2) ²=200²-2*2*200+2²=40000-800+4=39204
(7)(2a) ²•b^4=4a² b^4
(8) (- 2 / 3 A ^ 7b ^ 5) / / 3 / 2A & # 178; B ^ 5 = (- 2 * a ^ 7 B ^ 5 * 2A & # 178; B ^ 5) / (3 * 3) ---- a seven B five is not the denominator, right
= -4/9( a^9 b^10)
=(- 2 * 2A & # 178; B ^ 5) / (3 A ^ 7 B ^ 5 * 3) = - 4 / (9 A ^ 5) ---- a 7 b 5 is the denominator
(9)（6/5 a³ x^4-0.9ax³）÷3/5ax³=2a²x - 1.5
(10)（7x²y³ -8x³ y²z）÷8x²y²=(7/8) y - xz
(11)12^13÷(3^10•4^11)=(3*4)^13÷(3^10•4^11)=3³•4²=27•16=432
(12) (factorization)
1. 25X & # 178; - 16y & # 178; = (5x) & # 178; - (4Y) & # 178; = (5x-4y) (5x + 4Y) ---- square difference
2.(a-b)(x-y)-(b-a)(x+y)=(a-b)[(x-y)+(x+y)]=2x(a-b)
3.a²-4ab+4b²=a²- 2•a•2b+(2b) ²=(a-2b) ²
4. 4 + 12 (X-Y) + 9 (X-Y) &# 178; = 2 & # 178; + 2 * 2 * 3 (X-Y) + 9 (X-Y) &# 178; = [2 + 3 (X-Y)] &# 178; = (3x-3y + 2) &# 178; put away
.
Solve the equation x & # 178; - 8x + 1 = 0
x²-8x+1=0
x²-8x+16=15
(x-4)²=15
x-4=±√15
x=4±√15
If the fourth power of 3 A, the N + 2 power of B and the M-1 of 5 a times the 2n + 3 power of B are the same kind of terms, find the value of (M + n) (m-n)
Because they are of the same kind, M-1 = 4, M = 5, N + 2 = 2n + 3, n = - 1
(m+n)(m-n)=24
Some people say that no matter what real numbers x and Y take, the value of the algebraic formula X & # 178; + Y & # 178; - 12x + 8y + 53 must be positive. What's your opinion? Explain the reason
x²+y²-12x+8y+53
=x²-12x+36+y²+8y+16+1
=(x-6)²+(y+4)²+1
Because (X-6) &# 178; ≥ 0, (y + 4) &# 178; ≥ 0
So x & # 178; + Y & # 178; - 12x + 8y + 53 ≥ 1
That is, the value of X & # 178; + Y & # 178; - 12x + 8y + 53 must be positive
x²+y²-12x+8y+53
=x²-12x+36+y²+8y+16+1
=(x-6)²+(y+4)²+1
≥1
So no matter what real numbers x and Y take, the value of the algebraic formula X & # 178; + Y & # 178; - 12x + 8y + 53 must be positive
X & # 178; + Y & # 178; - 12x + 8y + 53 = x ^ 2-12x + 36 + y ^ 2 + 8y + 16 + 1 to (X-6) ^ 2 + (y-4) ^ 2 + 1
So it's greater than or equal to 1
x²+y²-12x+8y+53
=x²-12x+36+y²+8y+16+1
=(x-6)^2+(y+4)^2+1
The value of the algebraic formula X & # 178; + Y & # 178; - 12x + 8y + 53 must be positive
The solution equation is: X & # 178; - 8x = 9
x²-8x-9=0
(x+1)(x-9)=0
x=-1 9
X^2-8X=9,X^2-8X-9=0;
The factorization results in (X-9) * (x + 1) = 0;
X = 9 or - 1
x²-8x=9
x²-8x-9=0
（x-9）（X+1）=0
obtain. X = 9 or x = - 1
X1=9 X2=-1
x²-8x=9
x²-8x+16=9+16
(x-4)²=25
x-4=±5
∴x1=9, x2=-1
x²-8x=9
x²-8x-9=0
（x-9）（X+1）=0
The solution is: x = 9 or x = - 1
If the polynomial-4x (m power-1) times y (n power + 1) and 2x / 3 (2m power-3) times y (3N power-5) are similar terms, then M = n=
If the power of (2 m) = M-3 is to the power of (2 m-1)
It's the same kind
So M & # 178; - 1 = 2m & # 178; - 3
n²+1=3n²-5
Then M & # 178; = 2, n & # 178; = 3
therefore
m=±√2,n=±√3
m-1=2m-3
∴m=2
n+1=3n-5
N=3