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L(E) est l′algebre des endomorphisme continus d′un espace de Banach E, muni de la norme d′application lineaire ; GL(E) est le sous ensemble des elements inversibles de L(E) a)Montrer que GL(E) est ouvert dans L(E) b)montrer que l′application ∅:GL(E)→GL(E) u→u^(−1) =∅(u) est continu dans GL(E) c)montrer que ∅ est differentiable dans GL(E) dt calculer d∅

$${L}\left({E}\right)\:{est}\:{l}'{algebre}\:{des}\:{endomorphisme} \\ $$$${continus}\:{d}'{un}\:{espace}\:{de}\:{Banach}\:{E}, \\ $$$${muni}\:{de}\:{la}\:{norme}\:{d}'{application}\:{lineaire} \\ $$$$;\:{GL}\left({E}\right)\:{est}\:{le}\:{sous}\:{ensemble}\:{des} \\ $$$${elements}\:{inversibles}\:{de}\:{L}\left({E}\right) \\ $$$$\left.{a}\right){Montrer}\:{que}\:{GL}\left({E}\right)\:{est}\:{ouvert}\:{dans} \\ $$$${L}\left({E}\right) \\ $$$$\left.{b}\right){montrer}\:{que}\:{l}'{application}\: \\ $$$$\emptyset:{GL}\left({E}\right)\rightarrow{GL}\left({E}\right)\: \\ $$$$\:\:\:\:\:\:\:{u}\rightarrow{u}^{−\mathrm{1}} =\emptyset\left({u}\right)\:{est}\:{continu}\:{dans}\:{GL}\left({E}\right) \\ $$$$\left.{c}\right){montrer}\:{que}\:\emptyset\:{est}\:{differentiable} \\ $$$${dans}\:{GL}\left({E}\right)\:{dt}\:{calculer}\:{d}\emptyset \\ $$

Question Number 176472 Answers: 1 Comments: 0

solve for x,y,z with x+y+z=x^2 +y^2 +z^2 =x^3 +y^3 +z^3 =5

$${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$

Question Number 176468 Answers: 1 Comments: 1

((sin (2x+18°))/(sin (2x+12°))) =(√((sin 36°)/(sin 48°))) tan 2x = (√(tan M)) .(√(tan N)) 0°<M,N<90° ⇒M+N=?°

$$\:\:\frac{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{18}°\right)}{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{12}°\right)}\:=\sqrt{\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{sin}\:\mathrm{48}°}}\: \\ $$$$\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\sqrt{\mathrm{tan}\:{M}}\:.\sqrt{\mathrm{tan}\:{N}} \\ $$$$\:\:\mathrm{0}°<{M},{N}<\mathrm{90}°\:\Rightarrow{M}+{N}=?° \\ $$

Question Number 176462 Answers: 1 Comments: 1

In ΔABC given ((2a)/(tan A)) = (b/(tan B)) then ((sin^2 A−cos^2 B)/(cos^2 A+cos^2 B))=?

$${In}\:\Delta{ABC}\:{given}\:\frac{\mathrm{2}{a}}{\mathrm{tan}\:{A}}\:=\:\frac{{b}}{\mathrm{tan}\:{B}}\: \\ $$$$\:{then}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {A}−\mathrm{cos}\:^{\mathrm{2}} {B}}{\mathrm{cos}\:^{\mathrm{2}} {A}+\mathrm{cos}\:^{\mathrm{2}} {B}}=? \\ $$

Question Number 176458 Answers: 5 Comments: 0

x^2 +x=1 ((x^5 +8)/(x+1))=?

$${x}^{\mathrm{2}} +{x}=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}}=? \\ $$

Question Number 176453 Answers: 2 Comments: 0

If , α , β , γ ∈ ( 0 , 1 ) , then prove that : (√((1−^ α ).(1−^ β ). (1−^ γ ))) +(√(α^ .β^ .γ^ )) < 1

$$ \\ $$$$\:\:\:{If}\:,\:\:\alpha\:,\:\beta\:,\:\gamma\:\in\:\left(\:\mathrm{0}\:\:,\:\:\mathrm{1}\:\right)\:\:,\:\:{then}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:{prove}\:\:{that}\::\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\left(\mathrm{1}\overset{} {−}\alpha\:\right).\left(\mathrm{1}\overset{} {−}\beta\:\right).\:\left(\mathrm{1}\overset{} {−}\gamma\:\right)}\:+\sqrt{\overset{} {\alpha}.\overset{} {\beta}.\overset{} {\gamma}}\:\:<\:\mathrm{1} \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 176449 Answers: 0 Comments: 0

A fair dice was thrown twice and it landed on a and b respectively then the probability that cubic equation x^3 −(3a+1)x^2 +(3a+2b)x−2b=0 has three distinct root

$$\mathrm{A}\:\mathrm{fair}\:\mathrm{dice}\:\mathrm{was}\:\mathrm{thrown}\:\mathrm{twice}\:\mathrm{and} \\ $$$$\mathrm{it}\:\mathrm{landed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{respectively}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{cubic}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{3}} −\left(\mathrm{3a}+\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3a}+\mathrm{2b}\right)\mathrm{x}−\mathrm{2b}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{three}\:\mathrm{distinct}\:\mathrm{root} \\ $$$$ \\ $$

Question Number 176448 Answers: 0 Comments: 3

Suppose a^3 +b^3 +c^3 =a^2 +b^2 +c^2 =a+b+c Prove that abc=0

$$\mathrm{Suppose}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \\ $$$$=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{abc}=\mathrm{0} \\ $$$$ \\ $$

Question Number 176768 Answers: 0 Comments: 0