Відео

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Nice video

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How can you prove that there is no other solutions?

I saw the letter D pentated to the power of 3 in my dad’s car

(2^20)-4= 1,048,572

asnwer=5.3 isit

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f'(x) = [f(x+dx) - f(x)]/dx [1/(x+dx) - 1/x]/dx = [x - (x+dx)]/[x(x+dx)dx] = -dx/[(x^2 + xdx)dx] = -1/(x^2 +xdx) Take the limit as dx goes to zero: f'(x) = -1/x^2

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Thank you sir

I have harder f(x+70)=65x³+890x+2^x

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X=2

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So to make this easier, i could basicly turn it into a root. Starting with let's say 3. Then draw a few branches, and turn them into tetrations

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Can you solve √5x + 4^2x = 906

3^x . 2^(2(1-1/x)) = 3^2 . 2^1 you get two equations- x = 2 and 2-2/x = 1 both give x =2

seeing that is traumatic

If g((x-1)/3)=x+7, then x=3g+1, and x+7=3g+8. Then g(x+1)=3(x+1)+8= 3x+11

(x^2) - 3 = 0 x^2 = 3 x = +/- 3 The 2nd and 3rd root, Simpler and elegant, thanks.

why using ln? 3^x * 4^((x-1)/x) = 18 3^x * 2^(2 * (x-1)/x) = 3^2 * 2^1 compare by same bases: 1=> 3^x=3^2 ------------------> x=2 2 => 2 * (x-1)/x = 1 (2x -2)/x = 1 2x - 2 = 1 * x 2x - x = 2 ------------------> x=2 check the first question with that simple solution... what am i missing?

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sqrt(x.x.x)/sqrt(x+x+x)=1 Note that x can't be 0. Simplifyimg xsqrt(x)/sqrt(3x)=1 xsqrt(x)-sqrt(3x)=0 sqrt(x)[x-sqrt(3)]=0 sqrt(x)=0 --> x=0 rejected x-sqrt(3) --> x=sqrt(3)

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x^(3/2)/(√3√x) = 1 x = √3

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That's very nice sir.

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But -sqrt(3) does not satisfy the equation either. You'd have the square root of a negative number (for the smart*ss who's gonna complain about complex numbers, the square root is not a well defined function in that field)

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√x³ = √3x x³ = 3x => x²=3 x = ± √3

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Good master

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Keep it up big bro your are very good