A simple calculation problem in the second grade of junior high school: (A & # 178; - AB + B & # 178;) (A & # 178; + AB + B & # 178;) Calculation (A & # 178; - AB + B & # 178;) (A & # 178; + AB + B & # 178;) Ask detailed problem-solving process, thank you Xueba!

A simple calculation problem in the second grade of junior high school: (A & # 178; - AB + B & # 178;) (A & # 178; + AB + B & # 178;) Calculation (A & # 178; - AB + B & # 178;) (A & # 178; + AB + B & # 178;) Ask detailed problem-solving process, thank you Xueba!

（a²-ab+b²）（a²+ab+b²）
=（a²+b²）²-a²b²
=a^4+2a²b²+b^4-a²b²
=a^4+a²b²+b^4
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（a²-ab+b²）（a²+ab+b²）
＝[﹙a²＋b²﹚－ab][﹙a²＋b²﹚＋ab]
＝﹙a²＋b²﹚²－a²b²
＝a^4＋2a²b²＋b^4－a²b²
＝a^4＋a²b²＋b^4.
(a²-ab+b²)(a²+ab+b²)
=[(a²+b²)+ab][(a²+b²)-ab]
=(a²+b²)²-(ab)²
=a⁴+2a²b²+b⁴-a²b²
=a⁴+a²b²+b⁴
Solve the equation; the square of 4 [x minus 5] = 16
Square of 4 [x minus 5] = 16
2 square (X-5) square = 4 square
2（x-5)=±4
x-5=±2
x1=7
x2=3
In the first quarter of production, the output value of a factory in January is 2.5 million yuan, and the monthly growth rate of the output value in February and March is the same. It is known that the total output value in the first quarter is 8.436 million yuan, so the monthly growth rate of February and March is calculated
Suppose that the monthly growth rate of February and March is X. according to the meaning of the question, 250 + 250 (1 + x) + 250 (1 + x) 2 = 843.6250 + 250 + 250x + 250 + 500X + 250x2 = 843.6250x2 + 750x-93.6 = 0, (5x + 15.6) (50x-6) = 0, the solution is X1 = - 3.12 (negative value is not the meaning of the question, rounding off), X2 = 0.12 = 12%. Answer: the monthly growth rate of February and March is 12%
How to solve the quadratic equation of one variable with y square + 5Y + 2 = 0
Note: the formula is y = (- B ± √ (b ^ 2-4ac)) / 2A y ^ 2 + 5Y + 2 = 0 (y ^ 2 represents the square of Y), where a = 1, B = 5, C = 2, because B ^ 2-4ac = 25-8 = 17 > 0, y has two different real roots: Y1 and Y2Y = (- B ± √ (b ^ 2-4ac)) / 2A = (- 5 ± √ 17) / 2y1
This formula can be changed into the square of (Y's square + 5 / 2) - 25 / 4 + 2 = 0, and the solution is y = - (5 plus minus root sign 17) / 2
The formula method for solving quadratic equation of one variable a = 1, B = 5, C = 2, y = - B ± √ (b ^ 2-4ac)] / (2a)
y=-b±√(b^2-4ac)]/(2a)
Where a = 1, B = 5, C = 2
Just put it in
Where √ (b ^ 2-4ac) is the square root of B minus 4ac
Calculation questions (1) - X & # 178; · (- x) &# 178; (2) (X-2) (x + 3)
1. Calculation questions (1) - X & # 178; · (- x) & # 178; (2) (X-2) (x + 3) (3) (a-2b + 3C) (a + 2b-3c) (4) (a + 1 / 2) & # 178; · (A & # 178; + 1 / 4) & # 178; · (A-1 / 2) & # 178;
2. If X & # 178; + 2x + Y & # 178; - 6y + 10 = 0, find the value of X, y
1. Calculation
(1)－x²·﹙－x﹚²=-x^4
(2)﹙x－2﹚﹙x＋3﹚ = x²+x-6
(3)﹙a－2b＋3c﹚﹙a＋2b－3c﹚
=a² -(2b-3c)²
=a²-4b²+12bc-9c²
﹙4﹚﹙a＋1/2﹚²·﹙a²＋1/4﹚²·﹙a－1/2﹚²
=﹙a²-1/4﹚²·﹙a²＋1/4﹚²
=﹙a^4－1/16﹚²
=a^8-1/8a^4-1/256
2. If X & # 178; + 2x + Y & # 178; - 6y + 10 = 0,
Then (x + 1) ² + (Y-3) ² = 0
x+1=0 x=-1
y-3=0 y=3
Equation (x ^ 2-1) ^ 2-5 (x ^ 2-1) + 4 = 0. We can treat (x ^ 2-1) as a whole. Let x ^ 2-1 = y. then (x ^ 2-1) ^ 2 = y ^ 2, y ^ 2-5y + 4 = 0
The solution is Y1 = 1, y2 = 4, when y = 1 x ^ 2-1 = 1, x ^ 2 = 2, so x = positive and negative root sign 2, when y = 4, x ^ 2-1 = 4, so x ^ 2 = 5, so x + positive and negative root sign 5
X1 = root 2 x2 = - root 2 X3 = root 5 X4 = - root 5
This method is used to solve the problem that the fourth power of X - the second power of X - 6 = 0
Let x ^ 2 = y
The fourth power of X - the second power of X - 6 = 0 becomes
y^2-y-6=0
Y = 3 or y = - 2 (rounding off)
x^2=3
x=±√3
1 × 2 / 1 + 2 × 3 / 1 + 3 × 4 / 1 + 4 × 5 / 1
1 × 2 / 1 + 2 × 3 / 1 + 3 × 4 / 1 + 4 × 5 / 1
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5
=1-1/5
=4/5
The formula method is used to solve the following equation: 1.xsquare-x-1 = 0.2.5y-84 + ysquare = 0
Solve the following equation with proper method: 1. X square + 6x-5 = 0.2. (x + 3) (x-3) = 2
x²+6x-5=0
x²+6x+9=5+9
(x+3)²=14
x+3=±√14
x=-3±√14
2.（x+3)(x-3)=2
x²-9=2
x²=11
x=±√11
2 and 2 out of 9 - (2 out of 9 + 5 out of 12) 1, 8 out of 9-2 out of 3 + 1 out of 3
2 and 2 / 9 - (2 / 9 + 5 / 12)
=2 and 2 / 9-2 / 9-5 / 12
=2-5/12
=19/12
1 and 8 / 9-2 / 3 + 1 / 3
=1 and 8 / 9 - (2 / 3-1 / 3)
=1 and 8 / 9-1 / 3
=1 and 8 / 9-3 / 9
=1 and 5 / 9
Solve the equation: the square of (2x-3) - 4 (X-2) (x = 2) = 1, simplify the evaluation: the square of (x-5y) (- x-5y) - (- x + 5Y), where x = 0.5, y = - 1
Solving equation: the square of (2x-3) - 4 (X-2) (x = 2) = 1
Simplify the square of (x-5y) (- x-5y) - (- x + 5Y), where x = 0.5, y = - 1
Solving equation: the square of (2x-3) - 4 (X-2) (x + 2) = 1
-50y ^ 2A ^ 2 means square
（2x-3）^2-4（x-2）（x+2）=1 （x-5y）(-x-5y)-(-x+5y)^2
(4x^2-12x+9)-4(x^2-4)=1 = -25y^2+x^2-x^2+10xy-25y^2
4x^2-12x+9-4x^2+16=1 =10xy+ -50y^2
-12x + 9 + 16 = 1 substitute x = 0.5 y = - 1 into the above formula
x=2 =-55
Solving equation: the square of (2x-3) - 4 (X-2) (x = 2) = 1
？？？
The square of (2x-3) - 4 (X-2) (x + 2) = 1
=4X (square) + 9-12x-4 (x (square) - 4) = 1
Expansion = 4x (square) + 9-12x-4x (square) + 16 = 1
Eliminate 4x (square)
We get 9-12x + 16 = 1
X=2
Simplify the square of (x-5y) (- x-5y) - (- x + 5Y), where x = 0.5, y = - 1
(x-5y) (- x-5y) - (- x + 5Y) (- x + 5... Expansion)
The square of (2x-3) - 4 (X-2) (x + 2) = 1
=4X (square) + 9-12x-4 (x (square) - 4) = 1
Expansion = 4x (square) + 9-12x-4x (square) + 16 = 1
Eliminate 4x (square)
We get 9-12x + 16 = 1
X=2
Simplify the square of (x-5y) (- x-5y) - (- x + 5Y), where x = 0.5, y = - 1
（x-5y）(-x-5y)-(-x+5y)（-x+5y）=（x-5y）(-x-5y)+(x-5y)（-x+5y）
Common factor = (x-5y) (- x-5y-x + 5Y) = (x-5y) (- 2x) carry in value = - 5.5 ﹣ put away