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Question Number 121102 Answers: 2 Comments: 0

∫ (((x−1)(√(x^4 +2x^3 −x^2 +2x+1)))/(x^2 (x+1))) dx ?

$$\:\int\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)}\:\mathrm{dx}\:? \\ $$

Question Number 121100 Answers: 1 Comments: 0

Question Number 121099 Answers: 1 Comments: 0

Let α be a root of x^5 −x^3 +x−2=0 Then prove that [α^6 ]=3 where[λ] denotes greatest integer less than or equal λ

$${Let}\:\alpha\:{be}\:{a}\:{root}\:{of}\:\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$${Then}\:{prove}\:{that}\:\:\:\left[\alpha^{\mathrm{6}} \right]=\mathrm{3}\:\:\:\:\:\:\:{where}\left[\lambda\right]\:\:{denotes}\:{greatest}\:{integer} \\ $$$${less}\:{than}\:{or}\:\:{equal}\:\lambda \\ $$

Question Number 121097 Answers: 1 Comments: 0

Question Number 121092 Answers: 0 Comments: 0

Question Number 121091 Answers: 0 Comments: 1

Question Number 121085 Answers: 3 Comments: 1

Question Number 121081 Answers: 0 Comments: 1

Question Number 121077 Answers: 5 Comments: 1

Question Number 121074 Answers: 2 Comments: 0

lim_(x→1) ((x−1)/( (√(x^2 −1))))?

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}? \\ $$

Question Number 121073 Answers: 1 Comments: 0

Question Number 121082 Answers: 0 Comments: 0

Question Number 121066 Answers: 3 Comments: 1

1)∫(x/(x+1))dx=? 2)∫(x/(√(x−1)))dx=? 3)∫x(√(x−4))dx=?^

$$\left.\mathrm{1}\right)\int\frac{{x}}{{x}+\mathrm{1}}{dx}=? \\ $$$$\left.\mathrm{2}\right)\int\frac{{x}}{\sqrt{{x}−\mathrm{1}}}{dx}=? \\ $$$$\left.\mathrm{3}\right)\int{x}\sqrt{{x}−\mathrm{4}}{dx}=\overset{} {?} \\ $$$$ \\ $$

Question Number 121064 Answers: 0 Comments: 0

Question Number 121052 Answers: 3 Comments: 0

show that lim_(x→0) ((1−cosx)/x^2 )=(1/2)

$$\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 121051 Answers: 2 Comments: 0

show that lim_(x→0) ((tanx)/x)=1

$$\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tanx}}{\mathrm{x}}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 121050 Answers: 2 Comments: 0

show that lim_(x→0 ) ((sinx)/x)=1

$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\:\frac{\mathrm{sinx}}{\mathrm{x}}=\mathrm{1} \\ $$

Question Number 121046 Answers: 4 Comments: 0

1.∫e^x sinxdx=? 2.∫e^x cosxdx=?

$$\mathrm{1}.\int{e}^{{x}} {sinxdx}=? \\ $$$$ \\ $$$$\mathrm{2}.\int{e}^{{x}} {cosxdx}=? \\ $$

Question Number 121045 Answers: 0 Comments: 0

Question Number 121044 Answers: 0 Comments: 0

(3/4)+(1/2)=..... { (),() :}

$$\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}=.....\begin{cases}{}\\{}\end{cases} \\ $$

Question Number 121042 Answers: 2 Comments: 0

prove that lim_(x→∞) (1+(1/x))^x = e

$$\mathrm{prove}\:\mathrm{that}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} =\:\mathrm{e}\: \\ $$

Question Number 121041 Answers: 1 Comments: 0

Σ_(k=1) ^n (−1)^k .k =?

$$\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} .\mathrm{k}\:=?\: \\ $$

Question Number 121031 Answers: 0 Comments: 1

lim_(x→0) ((x+ ∣x∣ (1+x))/x).sin ((1/x))?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}+\:\mid\mathrm{x}\mid\:\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}.\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)? \\ $$

Question Number 121029 Answers: 2 Comments: 1

Question Number 121028 Answers: 1 Comments: 0

Question Number 121014 Answers: 1 Comments: 0

show by recurrence that ∀ n≥1 , a^n −b^n =(a−b)(a^(n−1) +a^(n−2) ∗b+...+ab^(n−2) +b^(n−1) )

$$\mathrm{show}\:\mathrm{by}\:\mathrm{recurrence}\:\mathrm{that} \\ $$$$\forall\:\mathrm{n}\geqslant\mathrm{1}\:, \\ $$$$\mathrm{a}^{\mathrm{n}} −\mathrm{b}^{\mathrm{n}} =\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}^{\mathrm{n}−\mathrm{1}} +\mathrm{a}^{\mathrm{n}−\mathrm{2}} \ast\mathrm{b}+...+\mathrm{ab}^{\mathrm{n}−\mathrm{2}} +\mathrm{b}^{\mathrm{n}−\mathrm{1}} \right) \\ $$

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