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Question Number 195017 Answers: 3 Comments: 0

(x+1)^3 =1 x=?

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 195015 Answers: 1 Comments: 0

Question Number 195013 Answers: 1 Comments: 0

lim_(x→3) (((2((√6)−(√(2x)) +(√(6−2x))))/( (√(36−4x^2 )))) )

$$\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2x}}\:+\sqrt{\mathrm{6}−\mathrm{2x}}\right)}{\:\sqrt{\mathrm{36}−\mathrm{4x}^{\mathrm{2}} }}\:\right) \\ $$

Question Number 195012 Answers: 2 Comments: 0

Question Number 195003 Answers: 1 Comments: 0

y=lnx^x^x y^′ =?

$${y}={lnx}^{{x}^{{x}} } \:\:\:\:\:\:\:\:\:{y}^{'} =? \\ $$

Question Number 195002 Answers: 1 Comments: 0

x^(√x) =((√x))^x x=?

$${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$

Question Number 194998 Answers: 0 Comments: 2

Resolution du probldme pose par sonukgindia (16.7.2023) voir Q194819 △ABC AM=AN=ADcos (𝛂/2) { ((AC=AM+MC=17 (1))),((AB=AN+NB =18 (2))) :} AB−AC=1=NB−MC (3) △CDE cos (C/2)=((CM)/(CD))=((CE)/(CD)) ⇒CM=CE △BDE cos (B/2)=((BE)/(BD))=((BN)/(BD))⇒BN=BE ⇒BE−CE=1 BC=BE+CE=2BE−1 △BEF BF=9 cos B=((BE)/9) BE=9cos B ⇒BC=18cos B−1 Posons BC=x x=18cos B−1 d apres triangle ABC AC^2 =AB^2 +BC^2 −2AB.BCcos B ⇒17^2 =18^2 +x^2 −36(((x+1)/(18))) x^2 +2x−35=0 alors x=5

Question Number 194985 Answers: 0 Comments: 0

lim_(n→+∞) [((1/n))^n +((2/n))^n +...(((n−1)/n))^n ]=?

$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\left[\left(\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{2}}{\mathrm{n}}\right)^{\mathrm{n}} +...\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} \right]=? \\ $$

Question Number 194991 Answers: 0 Comments: 4

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) Find (a_1 )

$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${Find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 194990 Answers: 0 Comments: 0

If (f(x))=x^2 −x+1 find the value of f(1971)+f(2021)+f(50)

$$\mathrm{I}{f}\:\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${f}\left(\mathrm{1971}\right)+{f}\left(\mathrm{2021}\right)+{f}\left(\mathrm{50}\right) \\ $$

Question Number 194988 Answers: 1 Comments: 0

ABC=CBF AB

$${ABC}={CBF} \\ $$$${AB} \\ $$

Question Number 194971 Answers: 1 Comments: 2

Question Number 194968 Answers: 2 Comments: 0

Question Number 194967 Answers: 1 Comments: 0

Question Number 194963 Answers: 2 Comments: 0

Question Number 194961 Answers: 1 Comments: 0

Question Number 194960 Answers: 1 Comments: 0

Soit x>1. On de^ finie la suite (p_n ) par p_1 =x et ∀n∈IN^∗ p_(n+1) =2p_n ^2 −1 Montrer que lim_(n→+∞) Π_(k=1) ^n (1+(1/p_k ))=(√((x+1)/(x−1)))

$$\mathrm{Soit}\:{x}>\mathrm{1}.\:\mathrm{On}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{la}\:\mathrm{suite}\:\left(\mathrm{p}_{\mathrm{n}} \right)\:\mathrm{par}\: \\ $$$$\mathrm{p}_{\mathrm{1}} ={x}\:\:\mathrm{et}\:\forall\mathrm{n}\in\mathrm{IN}^{\ast} \:\:\:\:\:\mathrm{p}_{\mathrm{n}+\mathrm{1}} =\mathrm{2p}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{Montrer}\:\mathrm{que}\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{p}_{\mathrm{k}} }\right)=\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}} \\ $$

Question Number 194975 Answers: 0 Comments: 6

∫(1/(x^5 +1))dx

$$\int\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{1}}\boldsymbol{{dx}} \\ $$

Question Number 194953 Answers: 1 Comments: 0

Let P(x)= x^2 +(x/2)+b and Q(x)=x^2 +cx+d be two polynomial with real coefficients such that P(x)Q(x)= Q(P(x)) for all real x . Find all the real roots of P(Q(x))=0

$$\:{Let}\:{P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:{and} \\ $$$$\:\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}\:{be}\:{two}\: \\ $$$$\:\:{polynomial}\:{with}\:{real}\:{coefficients} \\ $$$$\:\:{such}\:{that}\:{P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$$\:{for}\:{all}\:{real}\:{x}\:. \\ $$$$\:\:{Find}\:{all}\:{the}\:{real}\:{roots}\:{of}\: \\ $$$$\:\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\: \\ $$

Question Number 194952 Answers: 1 Comments: 0

x + (√(17−x^2 )) + x(√(17−x^2 )) =9 find the possible value of X

$$\boldsymbol{\mathrm{x}}\:+\:\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\:+\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{X}} \\ $$

Question Number 194942 Answers: 2 Comments: 2

= ((√5)+1)y + ((√5)+1)x x ≠ y and x+y ≠ 0 ((x/y))^(2025) +((y/x))^(2025)

$$\:\:\: =\: \left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\: +\:\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}\:\neq\:\mathrm{y}\:\mathrm{and}\:\mathrm{x}+\mathrm{y}\:\neq\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\: \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{2025}} +\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2025}} \\ $$

Question Number 194939 Answers: 0 Comments: 0