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Y = - (x + 1) the square of X + the square of root x + 2x + 1 to find the value of XY Please answer online Y = root sign - (x + 1) square of X + square of root sign x + 2x + 1 the problem of finding the value of XY has forgotten the root sign
Is it - (x + 1) x ^ 2 under the root
Given 2x + 2 - x = a (constant), find the value of 8x + 8 - X
Let 2x = t, then 2-x = T-1, then t + T-1 = a. ① method 1: T2 + T-2 = A2-2 from the square of both sides of ①, then 8x + 8-x = T3 + T-3 = (T + t-1) (t2-t · T-1 + T-2) = a (a2-3) = a3-3a. Method 2: 8x + 8-x = T3 + T-3 = (T + t-1) [(T + t-1) 2-3T · T-1] = a (a2-3) = a3-3a
Given x + y = 12, xy = 9, and x < y, find the value of (x ^ 1 / 2-y ^ 1 / 2) / (x ^ 1 / 2 + y ^ 1 / 2)
X < y (x ^ 1 / 2-y ^ 1 / 2) / (x ^ 1 / 2 + y ^ 1 / 2) < 0
The idea is square root
So: (x ^ 1 / 2-y ^ 1 / 2) / (x ^ 1 / 2 + y ^ 1 / 2)
=-{[(x^1/2-y^1/2)]^2/[(x^1/2+y^1/2)]^2}^1/2
=-{[x+y-2(xy)^1/2]/[x+y+2(xy)^1/2}^1/2
=-[(12-6)/(12+6)]^1/2
=-(1/3)^1/2
Given x + y = 12, xy = 9, find the value of x ^ 2 + y ^ 2
Such as the title
x^2+y^2
=x^2+y^2+2xy-2xy
=(x+y)^2-2xy
=12*12-2*9
=126
Given (x + y) ^ 2 = 9, (X-Y) ^ 2 = 5, find the value of XY
Solution: (x + y) ^ 2 = 9, (X-Y) ^ 2 = 5
Then x & # 178; + 2XY + Y & # 178; = 9... (1)
x²-2xy+y²=5.(2)
(1) - (2) get
4xy=4
xy=1
I wish you happiness and hope it will help you
Given x + y = 10, xy = 24, then 2x (2) + 2Y (2) =?
Two in parentheses denote the square of X and the square of Y
2X(2)+2Y(2)=2(X2+Y2)=2[(X+Y)2-2XY]=2*(100-48)=104
If x + y = 10, xy = 24, find the value of x ^ 2-y ^ 2
x^2-y^2=(x+y)(x-y)
(x-y)^2
=(x+y)^2-4xy
=100-96
=4
X-Y = - 2 or 2
x^2-y^2
=(x+y)(x-y)
=10×(±2)
=-20 or 20
Given x + y = 3, xy = - 10, then | X-Y|=
Because (X-Y) 2 = (x + y) 2-4xy
=9+40
=49
So | X-Y | = 7
x-y=3,xy=10,x^2+y^2=?
x^2+y^2
=(x-y)^2+2xy
=3^2+2*10
=29
Given X-Y = 3, xy = 10, then the value of (x + y) 2 is ()
A. 49B. 39C. 29D. 19
(x + y) 2 = (X-Y) 2 + 4xy = 9 + 40 = 49
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