Άρρητη εξίσωση 32
Δημοσιεύτηκε: Σάβ Αύγ 02, 2014 9:24 pm
Να λύσετε την εξίσωση ![8{x^3} + 8x - 4 = \sqrt[3]{{4 - 6x}}](/forum/ext/geomar/texintegr/latexrender/pictures/65449c55bc3013a65304cf1bbebb519c.png "8{x^3} + 8x - 4 = \sqrt[3]{{4 - 6x}}")
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Πηγή ασκήσεων 29-32 στην εικόνα
.Πηγή ασκήσεων 29-32 στην εικόνα
είναι ρίζα της εξίσωσης αφού την επαληθεύει !
Επίσης είναι και η μόνη, διότι η
![\displaystyle{
\,\,\,\,f(x) = \,\,8x^3 + \,\,8x\,\, - \,\,1 - \,\,\sqrt[3]{{4 - 6x\,}}\,\,\,\,\,\,\,\, \uparrow ( - \infty ,\frac{2}{3}]\,\,\,\,\,
}](/forum/ext/geomar/texintegr/latexrender/pictures/6cf3edcbc24fba6c0f18c3536698cbfc.png "\displaystyle{
\,\,\,\,f(x) = \,\,8x^3 + \,\,8x\,\, - \,\,1 - \,\,\sqrt[3]{{4 - 6x\,}}\,\,\,\,\,\,\,\, \uparrow ( - \infty ,\frac{2}{3}]\,\,\,\,\,
}")
(ή ενναλακτικά λόγω διάταξης !)![x \in \left( { - \infty ,\frac{2}{3}} \right]](/forum/ext/geomar/texintegr/latexrender/pictures/101e7ad8bf63fd86d81a87d223f23bcf.png "x \in \left( { - \infty ,\frac{2}{3}} \right]")
.![\displaystyle{8{x^3} + 8x - 4 = \sqrt[3]{{4 - 6x}} \Rightarrow }](/forum/ext/geomar/texintegr/latexrender/pictures/33b41d6cac35e292594903b77bf598f4.png "\displaystyle{8{x^3} + 8x - 4 = \sqrt[3]{{4 - 6x}} \Rightarrow }")
![\displaystyle{8{x^3} + 8x - 5 = \sqrt[3]{{4 - 6x}} - 1 \Rightarrow }](/forum/ext/geomar/texintegr/latexrender/pictures/23df52f41aa21c8f330f38784b82d42c.png "\displaystyle{8{x^3} + 8x - 5 = \sqrt[3]{{4 - 6x}} - 1 \Rightarrow }")
![\displaystyle{\left( {x - \frac{1}{2}} \right)\left( {8{x^2} + 4x + 10} \right) = \frac{{ - 6\left( {x - \frac{1}{2}} \right)}}{{{{\left( {\sqrt[3]{{4 - 6x}}} \right)}^2} + \sqrt[3]{{4 - 6x}} + 1}} \Rightarrow }](/forum/ext/geomar/texintegr/latexrender/pictures/8f2bd3ffb3ad19f17a7427ea06a9baa7.png "\displaystyle{\left( {x - \frac{1}{2}} \right)\left( {8{x^2} + 4x + 10} \right) = \frac{{ - 6\left( {x - \frac{1}{2}} \right)}}{{{{\left( {\sqrt[3]{{4 - 6x}}} \right)}^2} + \sqrt[3]{{4 - 6x}} + 1}} \Rightarrow }")
![\displaystyle{x = \frac{1}{2} \vee \underbrace {8{x^2} + 4x + 10}_ + = \underbrace {\frac{{ - 6}}{{{{\left( {\sqrt[3]{{4 - 6x}}} \right)}^2} + \sqrt[3]{{4 - 6x}} + 1}}}_ - \Rightarrow x = \frac{1}{2}}](/forum/ext/geomar/texintegr/latexrender/pictures/8b85f0a20e93b2c38049a30b185594dc.png "\displaystyle{x = \frac{1}{2} \vee \underbrace {8{x^2} + 4x + 10}_ + = \underbrace {\frac{{ - 6}}{{{{\left( {\sqrt[3]{{4 - 6x}}} \right)}^2} + \sqrt[3]{{4 - 6x}} + 1}}}_ - \Rightarrow x = \frac{1}{2}}")
είναι δεκτή.