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AlgebraQuestion and Answers: Page 91

Question Number 174735    Answers: 0   Comments: 0

Question Number 174732    Answers: 1   Comments: 0

Question Number 174719    Answers: 1   Comments: 0

Question Number 174709    Answers: 1   Comments: 3

Question Number 174684    Answers: 1   Comments: 0

Q: I , J are two ideals of commutative ring , ( R ,⊕, ) .prove that : (√( I ∩ J )) =^? (√( I )) ∩ (√( J )) m.n note : (√(I )) = { x ∈ R ∣ ∃ n∈ N , x^( n) ∈ I }

$$ \\ $$$$\boldsymbol{\mathrm{Q}}:\:\:\boldsymbol{\mathrm{I}}\:,\:\boldsymbol{\mathrm{J}}\:\:\boldsymbol{{are}}\:\boldsymbol{{two}}\:\boldsymbol{{ideals}}\:\boldsymbol{{of}}\:\:\boldsymbol{{commutative}}\: \\ $$$$\:\:\:\:\boldsymbol{{ring}}\:,\:\left(\:\boldsymbol{\mathrm{R}}\:,\oplus,\: \:\right)\:.\boldsymbol{{prove}}\:\boldsymbol{{that}}\:: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\:\boldsymbol{\mathrm{I}}\:\cap\:\boldsymbol{\mathrm{J}}\:}\:\:\overset{?} {=}\:\sqrt{\:\boldsymbol{\mathrm{I}}\:}\:\:\cap\:\:\sqrt{\:\boldsymbol{\mathrm{J}}\:}\:\:\:\:\:\boldsymbol{{m}}.\boldsymbol{{n}} \\ $$$$\:\:\:\:\boldsymbol{{note}}\::\:\sqrt{\boldsymbol{\mathrm{I}}\:}\:=\:\left\{\:\boldsymbol{{x}}\:\in\:\boldsymbol{\mathrm{R}}\:\mid\:\exists\:\boldsymbol{{n}}\in\:\mathbb{N}\:,\:\boldsymbol{{x}}^{\:\boldsymbol{{n}}} \:\in\:\boldsymbol{\mathrm{I}}\:\right\}\: \\ $$$$\: \\ $$

Question Number 174674    Answers: 0   Comments: 0

In △ABC the following relationship holds: a^4 < (b^2 + c^2 )^2 + 9R^4

$$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship}\:\mathrm{holds}: \\ $$$$\mathrm{a}^{\mathrm{4}} \:<\:\left(\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{9R}^{\mathrm{4}} \\ $$

Question Number 174673    Answers: 0   Comments: 0

Question Number 174668    Answers: 1   Comments: 0

Question Number 174655    Answers: 1   Comments: 0

Factorize a^3 − 16a − 3

$$\mathrm{Factorize}\:{a}^{\mathrm{3}} \:−\:\mathrm{16}{a}\:−\:\mathrm{3} \\ $$

Question Number 174654    Answers: 0   Comments: 1

Factorize x^9 + x^7 + 1

$$\mathrm{Factorize}\:{x}^{\mathrm{9}} \:+\:{x}^{\mathrm{7}} \:+\:\mathrm{1} \\ $$

Question Number 174649    Answers: 0   Comments: 3

Question Number 174647    Answers: 0   Comments: 0

Question Number 174635    Answers: 2   Comments: 0

Question Number 174587    Answers: 0   Comments: 2

Question Number 174579    Answers: 0   Comments: 0

find minimum value of: ((a^2 (b−1)^2 +2ab+(a−1)^2 )/(a(ab+1))) [all∈R]

$$\:\:\:\:{find}\:{minimum}\:{value}\:{of}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{b}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\boldsymbol{{ab}}+\left(\boldsymbol{{a}}−\mathrm{1}\right)^{\mathrm{2}} }{\boldsymbol{{a}}\left(\boldsymbol{{ab}}+\mathrm{1}\right)}\:\:\:\:\:\:\:\:\:\left[\boldsymbol{{all}}\in\boldsymbol{{R}}\right] \\ $$

Question Number 174569    Answers: 0   Comments: 0

{ ((x+z=0)),((y+t+xz=−2b)),((xt+yz=1)),((yt=b^2 −a)) :} [ all∈R] solve for : x, y ,z, t .

$$\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{0}}\\{\boldsymbol{{y}}+\boldsymbol{{t}}+\boldsymbol{{xz}}=−\mathrm{2}\boldsymbol{{b}}}\\{\boldsymbol{{xt}}+\boldsymbol{{yz}}=\mathrm{1}}\\{\boldsymbol{{yt}}=\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{a}}}\end{cases}\:\:\:\:\:\:\left[\:\:\boldsymbol{{all}}\in\boldsymbol{{R}}\right] \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{for}}\:\:\::\:\:\:\boldsymbol{{x}},\:\boldsymbol{{y}}\:,\boldsymbol{{z}},\:\boldsymbol{{t}}\:\:\:\:\:\:\:\:\:\:. \\ $$

Question Number 174568    Answers: 1   Comments: 0

If (1/a) + (1/b) + (1/c) = (1/(a + b + c)) then (1/a^3 ) + (1/(b^3 )) + (1/c^3 ) = ?

$$\mathrm{If}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:\mathrm{then}\: \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{3}} \:}\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{3}} }\:=\:? \\ $$

Question Number 174552    Answers: 2   Comments: 3

Question Number 174538    Answers: 1   Comments: 3

what is the maximum and minimum value of “x” if x+y+z = 6, x^2 +y^2 +z^2 = 18, for real values of “x,y and z”

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:``\mathrm{x}''\:\mathrm{if} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{6},\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{18}, \\ $$$$\mathrm{for}\:\mathrm{real}\:\mathrm{values}\:\mathrm{of}\:``\mathrm{x},\mathrm{y}\:\mathrm{and}\:\mathrm{z}'' \\ $$

Question Number 174533    Answers: 1   Comments: 0

Question Number 174532    Answers: 0   Comments: 0

Question Number 174512    Answers: 1   Comments: 0

how many integer a,b∈z^+ a^5 −b^5 =10(b+1)^2 −9

$${how}\:{many}\:{integer}\:{a},{b}\in{z}^{+} \\ $$$${a}^{\mathrm{5}} −{b}^{\mathrm{5}} =\mathrm{10}\left({b}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$$ \\ $$

Question Number 174487    Answers: 1   Comments: 0

Question Number 174479    Answers: 0   Comments: 1

Question Number 174464    Answers: 0   Comments: 0

In △ABC R∈(AB) , P∈(BC) , Q∈(CA) AR=3 , RB=1 , BP=6 , PC=2 , CQ=5 , QA=4 Prove that: PQ + QR + RP > ((21)/2)

$$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{R}\in\left(\mathrm{AB}\right)\:,\:\mathrm{P}\in\left(\mathrm{BC}\right)\:,\:\mathrm{Q}\in\left(\mathrm{CA}\right) \\ $$$$\mathrm{AR}=\mathrm{3}\:,\:\mathrm{RB}=\mathrm{1}\:,\:\mathrm{BP}=\mathrm{6}\:,\:\mathrm{PC}=\mathrm{2}\:,\:\mathrm{CQ}=\mathrm{5}\:,\:\mathrm{QA}=\mathrm{4} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{PQ}\:+\:\mathrm{QR}\:+\:\mathrm{RP}\:>\:\frac{\mathrm{21}}{\mathrm{2}} \\ $$

Question Number 174451    Answers: 0   Comments: 0

please help If x is directly porportional to z and y is also directly porportional to z then what is the value of xy propotional to ?

$${please}\:{help} \\ $$$${If}\:{x}\:{is}\:{directly}\:{porportional}\:{to}\:{z}\:\:{and}\: \\ $$$${y}\:{is}\:{also}\:{directly}\:{porportional}\:{to}\:{z} \\ $$$$\:{then}\:{what}\:{is} \\ $$$${the}\:{value}\:{of}\:\:{xy}\:{propotional}\:{to}\:\:? \\ $$$$ \\ $$

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