(2x^2+1)-(x^2-2x+1)

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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

simplify solve for inverse tangent line

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simplify (2x²+1)−(x²−2x+1)

Solution

x²+2x

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Solution steps

One step at a time

(2x²+1)−(x²−2x+1)

Apply rule: (a)=a

(2x²+1)=2x²+1

=2x²+1−(x²−2x+1)

Apply the distributive law: −(a−b)=−a+b

−(x²−2x+1)=−x²+2x−1

=2x²+1−x²+2x−1

Group like terms

=2x²−x²+2x+1−1

Add similar elements: 2x²−x²=x²

=x²+2x+1−1

Subtract the numbers: 1−1=0

=x²+2x

Solve instead

f(x)=(2x²+1)−(x²−2x+1)

ddx ((2x²+1)−(x²−2x+1))

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

(2x^2+1)-(x^2-2x+1)

Solution

Solution steps

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