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If the power of X - the power of y = 30, x + y = 6 / 5, find the value of X-Y Hurry up, it's not that I don't use my brain, but that I still don't have an eyebrow after using my brain You kind and smart people. Solve it for me
Square of X - square of y = (X-Y) * (x + y) = 30, and X + y = 6 / 5
So, X-Y = 25
If the x power of 2 = the y + 1 power of 8 and the Y power of 81 = the X-5 power of 9, then the Y power of x =?
The x power of 2 = the y + 1 power of 8, the x power of 2 = the 3 (y + 1) power of 2, x = 3 (y + 1) = 3Y + 3 ---------------- (1) the Y power of 81 = the X-5 power of 9, the 4Y power of 3 = the 2 (X-5) power of 3, 4Y = 2 (X-5) = 2x-10 ---------------- (2) substitute (1) into (2) 4Y = 2 (3Y + 3) - 104y = 6y + 6
The second power of (X-Y) - 2 (X-Y) + (X-Y) - 5 (X-Y)
The second power of (X-Y) - 2 (X-Y) + (X-Y) - 5 (X-Y)
=(x-y) ²- 2(x-y)+(x-y)-5(x-y)
=(x-y) ²- (x-y)-5(x-y)
=(x-y) ²- 6(x-y)
=(x-y)(x-y-6)
Find the quadratic power of (x + y) (X-Y) - (X-Y), where x = 5. Y = 2
The quadratic power of (x + y) (X-Y) - (X-Y)
=(x-y)(x+y-x+y)
=2y(x-y)
=2*2*(5-2)
=12
If the x power of 2 = the (y + 1) power of 4 and the Y power of 27 = the (x-1) power of 3, find the value of X-Y
Power X of 2 = power (y + 1) of 4 = > power X of 2 = power 2 (y + 1) of 2 = > x = 2 (y + 1)
Power y of 27 = power (x-1) of 3 = > power 3Y of 3 = power (x-1) of 3 = > power 3Y = X-1
x=4,y=1
x-y=3
Y = (1 + x ^ 2) to the power of X, find y '
y=(1+x^2)^x
Take logarithm on both sides, then
lny=ln(1+x^2)^x=x*ln(1+x^2)
therefore
y'/y=ln(1+x^2)+x/(1+x^2)*2x
=ln(1+x^2)+2x^2/(1+x^2)
y'=(1+x^2)^x*[ln(1+x^2)+2x^2/(1+x^2)]
I don't know how to simplify the following
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