Factorization square difference formula - x ^ 2 + y ^ 2

Factorization square difference formula - x ^ 2 + y ^ 2

The original formula = y ^ - x ^ = (Y-X) (y + x)

Factorization with square difference formula: x ^ 2 + 2Y + 1 =?

You're still in the form of + below, please. This is the factorization, the product
It should be (x + 1) (x + 1)

-(X-2) ^ 2 + (x + 1) ^ 2 factorization factor (square difference formula)

-(x-2)^2+(x+1)^2
=[(x+1)+(x-2)][(x+1)-(x-2)]
=3(2x-1)

Given 1 / X - 1 / y = 5, then the value of 2x + 5xy-2y / x-2xy-y is

1 / X-1 / y = 5, multiply both sides by XY, Y-X = 5xy
2x+5xy-2y=2x+y-x-2y=x-y
x-2xy-y=(x-y)(1+2/5)
2X+5XY-2Y/X-2XY-Y=(x-y)/(x-y)(1+2/5)=5/7

80% of a number is 1 / 2 more than 1 / 2 of 5 / 6. Find the number

55/48

It is known that 4x ^ 2-xy-3y ^ 2 = 0 is 2x / y. and X + 2Y / x-2y

4X ^ 2-xy-3y ^ 2 = 0 (4x + 3Y) (X-Y) = 04x + 3Y = 0 or X-Y = 0x = - 3 / 4 y or x = y when x = y, (x + 2Y) / (x-2y) = 3Y / (- y) = - 3 when x = - 3 / 4Y, (x + 2Y) / (x-2y) = (- 3 / 4Y + 2Y) / (- 3 / 4y-2y) = 5 / 4Y) / (- 11 / 4Y) = - 5 / 11

() minus 1 / 4 equals 1

Five quarters

In the triangle ABC, ∠ BAC is an obtuse angle, D is a point on the edge of BC, AC & # 178; = ad & # 178; + DC & # 178;, try to explain AB & # 178; - AC & # 178; = BD & # 178; - CD & # 178;

∵AC²=AD²+DC²
∴∠ADC=90°
That is, ADB = 90 degree
∴AB²=BD²+AD²
∴AB²-AC²=(BD²+AD²)-(AD²+DC²)=BD²-CD²

The price of a 400ml shampoo is 32 yuan, and that of a 200ml shampoo is 19 yuan. The supermarket has a shampoo promotion, 38 yuan per bottle,
Send a bottle of 19 yuan 200ml shampoo, Xiao Ming's mother takes 120 yuan to buy shampoo, how to buy the most cost-effective?
.

Does 38 yuan per bottle mean 400ml?
Calculate the price per 100ml
Buy 400ml 32 / 4 = 8
Buy 200ml 19 / 2 = 9.5
Buy 600ml 38 / (4 + 2) = 6.33 the cheapest
120-38 * 3 = 6:1800ml, 6 yuan left

Find the maximum and minimum values of the following functions in the specified range, z = XY, x ^ 2 + y ^ 2 ≤ 4
z=xy,{(x,y)|x^2+y^2≤4}

The solution is x ^ 2 + y ^ 2 ≤ 4
Let x = ksina, y = kcosa
So K ^ 2Sin ^ 2A + K ^ 2cos ^ 2A ≤ 4
That is, K ^ 2 ≤ 4
That is - 2 ≤ K ≤ 2
Then z = xy = ksinakcosa = k ^ 2 * 1 / 2 × 2sinacosa
=1/2k^2sin2a
From - 1 ≤ sin2a ≤ 1
That is - 1 / 2 ≤ 1 / 2sin2a ≤ 1 / 2
And K ^ 2 ≤ 4
That is - 1 / 2K ^ 2 ≤ 1 / 2K ^ 2sin2a ≤ 1 / 2K ^ 2
That is - 2 ≤ 1 / 2K ^ 2sin2a ≤ 2
So - 2 ≤ Z ≤ 2
So the maximum 2 and minimum - 2 of the function z = XY on x ^ 2 + y ^ 2 = 1