Simplify as much as possible.assume that all variables

Ответ:

christylam1606

31.12.2021

$4x\sqrt{32xv^2}+v\sqrt{2x^3}=4x\sqrt{16\cdot2}\cdot\sqrt{x}\cdot\sqrt{v^2}+v\sqrt2\cdot\sqrt{x^2\cdot x}\\\\=4x\cdot4\sqrt2\cdot\sqrt{x}\cdot v+v\sqrt2\cdot x\sqrt{x}=16xv\sqrt{2x}+xv\sqrt{2x}=17\sqrt{2x}\\\\========================================\\\\D;\\x\in < 0;\ \infty)\ \wedge\ v\in\mathbb{R}$

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Ответ:

kseniyayakimov

10.07.2020

Mathematics

$\cos^4 x - \sin^4 x = \cos 2x$ for the right-hand side, use cosine double angle identity which states that cos 2x = cos²x - sin² x $\text{rhs} = \cos 2x = \cos^2 x - \sin^2 x$ for the left-hand side $\text{lhs} = \cos^4 x - \sin^4 x$ we factor using difference of squares.recall difference of squares factoring: a² - b² = (a - b)(a + b)but here we have a⁴ - b⁴, so a⁴ - b⁴ = (a² - b²)(a² + b²)note that cos⁴ x is equivalent to (cos x)⁴ and etc. $\begin{aligned} \text{lhs} & = \cos^4 x - \sin^4 x \\ & = (\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) \end{aligned}$ note that cos²x + sin²x = 1 via the pythagorean identity $\begin{aligned} \text{lhs} & = \boxed{\stackrel{\text{equals to 1} }{(\cos^2 x + \sin^2 x)} }(\cos^2 x - \sin^2 x) \\ & = 1 (\cos^2 - \sin^2 x ) \\ & = \text{rhs} \end{aligned}$

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Simplify as much as possible.
assume that all variables represent positive real number
4x√32xv^2 + v√2x^3

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Simplify as much as possible.assume that all variables represent positive real number4x√32xv^2 + v√2x^3

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Simplify as much as possible.
assume that all variables represent positive real number
4x√32xv^2 + v√2x^3