directrice (y-3)^2=8(x-5)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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simplifier e^2ln(x)

Solution

x²

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Solution steps

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[

xy=12

12x²+y²=156

]

Isoler x pour xy=12: x=12y ; y≠ 0

Intégrer les solutions x=12y dans 12x²+y²=156

Pour 12x²+y²=156, remplacer x par 12y : y=12, y=−12, y=2√3, y=−2√3

Intégrer les solutions y=12, y=−12, y=2√3, y=−2√3 dans xy=12

Pour xy=12, remplacer y par 12: x=1

Pour xy=12, remplacer y par −12: x=−1

Pour xy=12, remplacer y par 2√3: x=2√3

Pour xy=12, remplacer y par −2√3: x=−2√3

Vérifier des solutions en les intégrant dans les équations d'origine

Par conséquent, les solutions finales pour xy=12, 12x²+y²=156 sont

(

`x`=1,	`y`=12
`x`=−1,	`y`=−12
`x`=2√3,	`y`=2√3
`x`=−2√3,	`y`=−2√3

)

Solve instead

f(x)=e^2ln(x)

ddx (e^2ln(x))

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

directrice (y-3)^2=8(x-5)

Solution

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