Matematik

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07. april 2019 af Christinaei (Slettet) - Niveau: B-niveau

Jeg sidder med en matematik opgave som jeg slet ikke kan fonde ud af.... jeg har vedhæftet billede af opgaverne....

Vedhæftet fil: EADB4416-03C1-498B-BCA2-663FE9D82C96.png

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Svar #1
07. april 2019 af mathon

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Svar #2
07. april 2019 af mathon

a)
$\frac{x}{(x-3)}-\frac{15}{16}=\frac{x^2}{(x-3)(x+3)}\qquad x\notin\{-3,3\}\qquad \textup{FN: }16(x-3)(x+3)$

Der multipliceres med fælesnævneren:

$\small 16(x+3)\cdot x-(x^2-9)\cdot 15=16\cdot x^2$

$\small 16x^2+48x-15x^2+135=16\cdot x^2$

$\small 15x^2-48x-135=0$

$\small 5x^2-16x-45=0$

$x=\left\{\begin{matrix} -\frac{9}{5}\\5 \end{matrix}\right.$

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Svar #3
07. april 2019 af mathon

b)
$\small \begin{array}{lllll} &x^2+y^2=9\\ &x\cdot y=2\Leftrightarrow y=\frac{2}{x} \\ \textup{hvoraf:}&x^2+\left ( \frac{2}{x} \right )^2=9&&x\neq0\\\\ &x^2+\frac{4}{x^2}=9\\\\ &x^4+4=9x^2\\\\ &x^4-9x^2+4=0\\\\ &x^2=\left\{\begin{matrix} \frac{9-\sqrt{65}}{2}\\ \frac{9+\sqrt{65}}{2} \end{matrix}\right.\\\\ &x=\left\{\begin{matrix} \sqrt{\frac{9-\sqrt{65}}{2}}\\\sqrt{\frac{9+\sqrt{65}}{2}} \end{matrix}\right. \end{array}$

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Svar #4
07. april 2019 af mathon

$\small \begin{array}{lllll} \sqrt{5x+7}-\sqrt{x+4}=\sqrt{2x-5}\qquad x\geq 2.5\\\\ \left (\sqrt{5x+7}-\sqrt{x+4} \right )^2= 2x-5 \\\\ 5x+7-2\sqrt{5x+7}\cdot \sqrt{x+4}+x+4=2x-5\\\\ 6x+11-2\sqrt{5x+7}\cdot \sqrt{x+4}=2x-5\\\\ 4x+16=2\sqrt{5x+7}\cdot \sqrt{x+4}\\\\ 2x+8= \sqrt{5x+7}\cdot \sqrt{x+4}\\\\ \left (2x+8 \right )^2=\left ( \sqrt{5x+7}\cdot \sqrt{x+4} \right )^2\\\\ 4x^2+32x+64=(5x+7)(x+4)\\\\ x^2-5x-36=0\\\\ x=\left\{\begin{matrix} -4\\9 \end{matrix}\right. \end{array}$

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Svar #5
07. april 2019 af oppenede

#4 Kvadreringen introducerer en ugyldig løsning som du ikke eksluderer!

f)
$\\\begin{align*} \sqrt{6x-5}&-\sqrt{2x-1}&=\ &8 \\ (y+8)&-y&=\ &8,\ y\geq0 \end{align*} \\ (y+8)^2=6x-5\ \land\ y^2=2x-1\ \land\ y\geq 0 \\[0.3cm] y^2+64+16y=6x-5\ \land\ y^2=2x-1\ \land\ y\geq 0 \\[0.3cm] 2x+68+16y=6x\ \land\ y^2=2x-1\ \land\ y\geq 0 \\[0.3cm] 17+4y=x\ \land\ y^2=2x-1\ \land\ y\geq 0 \\[0.3cm] 17+4y=x\ \land\ y^2=33+8y\ \land\ y\geq 0 \\[0.3cm] 17+4y=x\ \land\ y\in\{11,-3\}\ \land\ y\geq 0 \\[0.3cm] 17+4y=x\ \land\ y=11$

Dvs. x = 61.

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Svar #6
07. april 2019 af mathon

forgelemmelse:

$\small \small \small \begin{array}{lllll} \sqrt{5x+7}-\sqrt{x+4}=\sqrt{2x-5}\qquad {\color{Red} \mathbf{x\geq 2.5}}\\\\ \left (\sqrt{5x+7}-\sqrt{x+4} \right )^2= 2x-5 \\\\ 5x+7-2\sqrt{5x+7}\cdot \sqrt{x+4}+x+4=2x-5\\\\ 6x+11-2\sqrt{5x+7}\cdot \sqrt{x+4}=2x-5\\\\ 4x+16=2\sqrt{5x+7}\cdot \sqrt{x+4}\\\\ 2x+8= \sqrt{5x+7}\cdot \sqrt{x+4}\\\\ \left (2x+8 \right )^2=\left ( \sqrt{5x+7}\cdot \sqrt{x+4} \right )^2\\\\ 4x^2+32x+64=(5x+7)(x+4)\\\\ x^2-5x-36=0\qquad x\geq 2.5 \\\\ x=\left\{\begin{array}{lll} -4&\textup{som m\aa \ forkastes}\\9 \end{array}\right. \\\\ x=9 \end{array}$

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Svar #7
07. april 2019 af mathon

k)
$\small \begin{array}{llll} x+y=2\Leftrightarrow y=(2-x) \\ 2x^2+2(2-x)^2+3x\cdot (2-x)=43\\\\ 2x^2+2(4-4x+x^2)+6x-3x^2=43\\\\ 2x^2+8-8x+2x^2+6x-3x^2=43\\\\ x^2-2x-35=0\\\\ x=\left\{\begin{matrix} -5\\ 7 \end{matrix}\right. \end{array}$

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