範囲 20((x-3)(x+4))/((x-3)^2(x-5))

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

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級数解法 4xy^{′ ′}+2y^′+y=0

Solution

y=c₁√x(1−x6 +x²120 +… +(−1)ⁿxⁿ(2)(3)(4)(5)…(2n)(2n+1) +… )+c₂(1−x2 +x²24 +… +(−1)ⁿxⁿ(1)(2)(3)(4)…(2n−1)(2n) +… )

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逆関数の定義

y=f(x), x=g(y) であれば,関数 g は関数 f の逆である

y=20· (x−3)(x+4)(x−3)²(x−5)

x を以下で置き換える: y

x=20· (y−3)(y+4)(y−3)²(y−5)

yを解く x=20(y−3)(y+4)(y−3)²(y−5)

4x+10+√x²+160x+100x , 4x+10−√x²+160x+100x

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こんにちは, について質問する級数解法 4xy^{′ ′}+2y^′+y=0 または以下のオプションから選択してください

関数の逆関数を見つける方法を説明する

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

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範囲 20((x-3)(x+4))/((x-3)^2(x-5))

Solution

Solution steps

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