Use factorization to solve the following equation! ①﹙2x+3﹚²-2x-3=0 ②﹙2x-1﹚²-x²=0 ③2y﹙2+y﹚=-﹙y+2﹚ ④﹙y-1﹚²+2y﹙y-1﹚=0

Use factorization to solve the following equation! ①﹙2x+3﹚²-2x-3=0 ②﹙2x-1﹚²-x²=0 ③2y﹙2+y﹚=-﹙y+2﹚ ④﹙y-1﹚²+2y﹙y-1﹚=0

1 (2x + 3) & # 178; - 2x-3 = 0 solution (2x + 3) & # 178; = 2x + 3 2x + 3 = 0 or 2x + 3 = 1, then x = - 3 / 2 or x = - 12 (2x-1) & # 178; - X & # 178; = 0 solution square difference (2x-1-x) (2x-1 + x) = 0, that is, (x-1) (3x-1) = 0, then x = 1 or x = 1 / 33 2Y (2 + y) = - (y + 2) solution shift term 2
①﹙2x+3﹚²-2x-3=(2x+3)²-(2x+3)=(2x+3)*(2x+3-1)=(2x+3)*(2x+2)=0，x1=-3/2;x2=-1。
② (2x-1) &# 178; - X & # 178; = 4x & # 178; - 4x + 1-x & # 178; = 3x & # 178; - 4x + 1 = 3x & # 178; - 3x-x + 1 = 3x (x-1) - (x-1) = (x-1) * (3x-1) = 0, X1 = 1; x2... Expand
①﹙2x+3﹚²-2x-3=(2x+3)²-(2x+3)=(2x+3)*(2x+3-1)=(2x+3)*(2x+2)=0，x1=-3/2;x2=-1。
②﹙2x-1﹚²-x²=4x²-4x+1-x²=3x²-4x+1=3x²-3x-x+1=3x(x-1)-(x-1)=(x-1)*(3x-1)=0，x1=1;x2=1/3。
③2y﹙2+y﹚=-﹙y+2﹚，2y(2+y)+(y+2)=(y+2)*(2y+1)=0，y1=-2;y2=-1/2。
④﹙y-1﹚²+2y﹙y-1﹚=(y-1)*(y-1+2y)=(y-1)*(3y-1)=0；y1=1;y2=1/3。 Put it away
（1）（2x—1）^2=25(x+1)^2
（2）（3x+1）^2—3（3x+1）=4
（1）（2x—1）^2=25(x+1)^22x—1=±5(x+1)2x—1=5(x+1)5x-2x=-1-5x=-22x—1=-5(x+1)2x+5x=-5+17x=-4x=-4/7（2）（3x+1）^2—3（3x+1）=4（3x+1）^2—3（3x+1）-4=0(3x+1-4)(3x+1+1)=0(3x-3)(3x+2)=03x-3=0x=13x+2=0...
（1）（2x—1）^2=25(x+1)^2
2x-1 = 5x + 5 or 2x-1 = - 5x-5
X = 2 or x = - 4 / 7
（2）（3x+1）^2—3（3x+1）=4
(3x+1-4)(3x+1+1)=0
(3x-3)(3x+2)=0
X = 1 or x = - 2 / 3
1)（2x—1）^2-25(x+1)^2=0
==>(2x-1-5x-5)(2x-1+5x+5)=0
==>(-3x-6)(7x+4)=0
X = - 2 or x = - 4 / 7
2)（3x+1）^2—3（3x+1）-4=0
==>((3x+1)-4)((3x+1)+1)=0
==>(3x-3)(3x+2)=0
X = 1 or x = - 2 / 3
Question 1: (2x-1) ^ 2 = 25 (x + 1) ^ 2
4x^2-4x+1=25x^2+50x+25
21x^2+54x+24=0
7x^2+18x+8=0
(7x+4)(x+2)=0
x1=-4/7 x2=-2
Chapter... Unfold
Question 1: (2x-1) ^ 2 = 25 (x + 1) ^ 2
4x^2-4x+1=25x^2+50x+25
21x^2+54x+24=0
7x^2+18x+8=0
(7x+4)(x+2)=0
x1=-4/7 x2=-2
Second question:
（3x+1）^2—3（3x+1）=4
9x^2+6x+1-9x-3=4
9x^2-3x-6=0
3x^2-x-2=0
(3x+2)(x-1)=0
X1 = - 2 / 3, X2 = 1 / 2
The following equation 2 (x-3) &# 178; + 2 (X & # 178; - 1) = 4x + 1 is solved by factorization
2（x-3）²+2（x²-1）=4x+1
2(x²-6x+9)+2x²-2=4x+1
2x²-12x+18+2x²-3-4x=0
4x²-16x+15=0
(2x-3)(2x-5)=0
2X = 3 or 2x = 5
That is: x = 3 / 2 or x = 5 / 2
Please explain how to find the root of equation X3 + x2-8x-8 = 0 (accurate to 0.01)
Find the root of equation X3 + x2-8x-8 = 0 (accurate to 0.01)
x3+x2-8x-8=0
x2(x+1)-8(x+1)=0
(x2-8)(x+1)=0
x1=2.83,x2= -2.83,x3=-1
Change to x ^ 2 (x + 1) - 8 (x + 1) = (x + 1) (x ^ 2-8) = 0
The limit of limx - > + ∞ x [(√ X & # 178; + 1) - x]
limx[(√x^2+1)-x=limx[(√x^2+1)-x]*[(√x^2+1)+x]/[(√x^2+1)+x]
x→+∞ x→+∞
=limx/[(√x^2+1)+x]
x→+∞
=limx*(1/x)/{[(√x^2+1)+x]*(1/x)}
x→+∞
=lim1/{[√1+(1/x^2)]+1}
x→+∞
=1/(1+1)
=1/2
limx->+∞ x[(√x²+1)-x]=
limx->+∞ [x*(x²+1-x²)/((√x²+1)+x)]=
limx->+∞ x/(√x²+1)+x)=1/2
It is known that the lengths of AB and AC on both sides of △ ABC are the two real roots of the quadratic equation X2 - (2k + 3) x + K2 + 3K + 2 = 0 with respect to x, and the length of the third side is 5. (1) try to explain that the equation must have two unequal real roots; (2) when k is the value, △ ABC is a right triangle with BC as the hypotenuse; (3) when k is the value, △ ABC is an isosceles triangle, and find the perimeter of △ ABC
(1) It is proved that: ∵ (?) = (2k + 3) 2-4 (K2 + 3K + 2) = 1, ∵ (?) > 0, ∵ no matter what the value of K is, the equation always has two unequal real roots; (2) when △ ABC is a right triangle with BC as the hypotenuse, there are AB2 + ac2 = BC2 and ∵ BC = 5, the lengths of AB and AC on both sides are two real roots of the univariate quadratic equation X2 - (2k + 3) x + K2 + 3K + 2 = 0 of X. ∵ AB2 + ac2 = 25, AB + AC = 2K + 3, ab · AC =(AB + AC) 2-2-2aab, and (AB + AC) 2-2aab (AB + AC) 2-2aab AC = 25 {(2k + 3) 2-2 · (K2 + 3K + 2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-2) (K-5) (AB + AC) 2-2aab + AC = 2K + AC = 2K + AC = 2K + AC = 2K + AC = 2K + 3 (2k + 3) 2 (2k + AC = AB = AB = AB = 5, ab = 5, ab = AB = 5, ab = AB = AB = 5, ab = AB = 3, 5AC = K2 + 3K + 2, the solution is K =3 or 4, AC = 4 or 6, and the circumference of ABC is 14 or 16
Several factorization problems in grade two of junior high school
(1)X^3-1 (2) 2a^2+4ab+2b^2 (3) (XY)^-1 （4） （x-y)^2+4xy (5) 2x(a-2)-y(2-a)
1. X & sup3; - 1 = (x-1) (X & sup2; + X + 1) cubic difference formula 2.2A ^ 2 + 4AB + 2B ^ 2 = 2 (A & sup2; + 2Ab + B & sup2;) = 2 (a + b) & sup2; 3. (XY) & sup2; - 1 = (XY + 1) (XY-1) 4. (X-Y) ^ 2 + 4xy = x & sup2; - 2XY + Y & sup2; + 4xy = x & sup2; + 2XY + Y & sup2; = (x + y) & sup2; 5
1 / 2x (- 3x & # 178; + 2x-1) - 1 / 3x & # 178; (2x-6x & # 178;) Note: X is an unknown number, not a multiplier sign
Original = - 3 / 2x & # 179; + X & # 178; - 1 / 2x-2 / 3x & # 179; + 2x ^ 4
=2x^4-13/6x³+x²-1/2x
3/2x³+x²-1/2x-2/3x³+2x^4
=2x^4-13/6x³+x²-1/2x
The relation between the root and coefficient of quadratic equation with one variable is discussed
Solve the equation and find the sum and product of the following two equations
-2x^2+3=0
Because the sum of two = - B / A, the product of two = C / A
-If 2x ^ 2 + 3 = 0, a = - 2, B = 0, C = 3,
So the sum of the two is 0, and the product of the two is - 1.5
1. Verification: 11 to the 10th power - 1 can be divided by 100
2. It is known that the lengths of three sides of a triangle are a, B, C, and satisfy the following conditions: the square of a + the square of B + the square of C - AB BC AC = 0
1. I didn't know how to tell you at first. I'm afraid you don't understand it. Later, I thought about the decomposition factor. As a junior high school student, I should remember it. It's very helpful for future study! A ^ 2-1 = (A-1) (a + 1) a ^ 3-1 = (A-1) (a ^ 2 + A + 1) this two or three times decomposition factor should be. At this time, it can be analogical! You should remember it for the time being
1. Use the square difference formula: 11 ^ 10-1 = (11 ^ 5) ^ 2-1 = (11 ^ 5 + 1) (11 ^ 5-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 ^ 3-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 + 1) (11-1) = 10 * (11 ^ 5 + 1) (11 ^ 3 + 1) (11 ^ 3-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 + 1)
So it's divisible by 10
2. Formula: A ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC
... unfold
1. Use the square difference formula: 11 ^ 10-1 = (11 ^ 5) ^ 2-1 = (11 ^ 5 + 1) (11 ^ 5-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 ^ 3-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 + 1) (11-1) = 10 * (11 ^ 5 + 1) (11 ^ 3 + 1) (11 ^ 3-1) = (11 ^ 5 + 1) (11 ^ 3 + 1) (11 + 1)
So it's divisible by 10
2. Formula: A ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC
2(a^2+b^2+c^2)=2(ab+ac+bc)
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0
(a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0 if and only if a = b = C, so it is an equilateral triangle
11^10-1
=(11^5-1)*(11^5+1)
=11^5+1)(11^3+1)(11+1)
1.（10-1）10=10 10-10
2. Square of a + square of B + square of C - AB BC AC
(- AB BC AC, re merged)
=0
A = B or B = C or a = b = C....
In short, it's isosceles