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points of on same line (2, 110), (3, 165), (4, 220), (5, 275)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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∫

ddx

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

simplify solve for inverse tangent line

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展開する (2x+3)²−(3y+2)²

Solution

4x²+12x−9y²−12y+5

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Solution steps

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lim_{x→ ∞}(x⁴−8x²−10x²−3x−10 )

最高の分母べき乗で割る:x²−8−10x² 1−3x −10x²

=lim_{x→ ∞}(x²−8−10x² 1−3x −10x² )

lim_x→a[f(x)g(x) ]=lim_x→af(x)lim_x→ag(x) , lim_x→ag(x)≠0
不定形式の場合を除く

=lim_{x→ ∞}(x²−8−10x² )lim_{x→ ∞}(1−3x −10x² )

lim_{x→ ∞}(x²−8−10x² )=∞

lim_{x→ ∞}(1−3x −10x² )=1

=∞ 1

無限大の特性を適用する:∞c =∞

=∞

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こんにちは, について質問する展開する (2x+3)²−(3y+2)²

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

points of on same line (2, 110), (3, 165), (4, 220), (5, 275)

Solution

Solution steps

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