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3 years ago

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Solve the nonlinear inequality (x^2 -2x +3) / (x+1) less than or equal to 1

1 answer:

kondor19780726 [428]3 years ago

4 0

$x\not=-1\\\\
\dfrac{x^2-2x+3}{x+1}\leq1\\
\dfrac{x^2-2x+3}{x+1}-1\leq0\\
\dfrac{x^2-2x+3}{x+1}-\dfrac{x+1}{x+1}\leq0\\
\dfrac{x^2-3x+2}{x+1}\leq0\\
(x^2-3x+2)(x+1)\leq0\\
(x^2-x-2x+2)(x+1)\leq0\\
(x(x-1)-2(x-1))(x+1)\leq0\\
(x-2)(x-1)(x+1)\leq0\\
\boxed{x\in (\infty,-1\rangle\cup(1,2\rangle}$

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2 years ago

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IrinaK [193]

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Step-by-step explanation:

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3/12 can be reduced to 1/4

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3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

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$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

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$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

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3 years ago

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Liono4ka [1.6K]

Answer

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Step-by-step explanation:

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