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

Solutions > Radicals Multiply Calculator >

линейный xy^{\prime \prime}+y^{\prime}-y=0

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

☐²

x^☐

√☐

^□√☐

□□

log_□

π

θ

∞

∫

ddx

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

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Generated by AI

периодичность tan(2x−5)

Solution

π2

Decimal

1.57079…

Show Steps

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Solution steps

One step at a time

(

1	2	1
6	−1	0
−1	−2	−1

)

Собственные векторы матрицы

Пусть A будет квадратной матрицей, η вектором и λ скаляром, которые удовлетворяют Aη=λη,
тогда η является собственным вектором A, а λ является собственным значением, связанным с ним

тобы найти собственные векторы η, решите (A−λ I)η=0 для каждого собственного значения λ

Найдите собственные значения для (

1	2	1
6	−1	0
−1	−2	−1

): λ=0, λ=3, λ=−4

Собственные векторы для λ=0: (

−1

−6

13

)

Собственные векторы для λ=3: (

−2

−3

2

)

Собственные векторы для λ=−4: (

−1

2

1

)

Собственные векторы для (

1	2	1
6	−1	0
−1	−2	−1

)

=(

−1

−6

13

), (

−2

−3

2

), (

−1

2

1

)

Solve instead

домен tan(2x−5)

диапазон tan(2x−5)

амплитуда tan(2x−5)

Radicals Multiply Examples

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