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AlgebraQuestion and Answers: Page 363

Question Number 15927 Answers: 0 Comments: 5

Question Number 15916 Answers: 2 Comments: 2

Question Number 15869 Answers: 1 Comments: 1

If ax^2 +(b/x)≥c for all x,a,b>0 then prove minimum value of 27ab^2 is 4c^2 .

$$\mathrm{If}\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant{c}\:\mathrm{for}\:\mathrm{all}\:{x},{a},{b}>\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{27}{ab}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{4}{c}^{\mathrm{2}} . \\ $$

Question Number 15865 Answers: 2 Comments: 0

a,b,c∈R^(+ ) andIf a+b+c=18 then maximum value of a^2 b^3 c^4 is

$${a},{b},{c}\in\mathbb{R}^{+\:} \mathrm{andIf}\:{a}+{b}+{c}=\mathrm{18}\:\mathrm{then}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} \:\mathrm{is} \\ $$

Question Number 15721 Answers: 1 Comments: 0

If x+y+z=1 with 0<x, y, z <(1/2) then find tbe range of values of (1/(x+y))+(1/(y+z))+(1/(z+x)) .

$${If}\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:{with}\:\mathrm{0}<{x},\:{y},\:{z}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{then}\:{find}\:{tbe}\:\:{range}\:{of}\:{values}\:{of} \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:. \\ $$

Question Number 15656 Answers: 1 Comments: 0

Solve : 0 ≤ x^2 − 5x + 7 < 1

$$\mathrm{Solve}\::\:\mathrm{0}\:\leqslant\:{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{7}\:<\:\mathrm{1} \\ $$

Question Number 15570 Answers: 2 Comments: 1

Solve: 2^x = x^4

$$\mathrm{Solve}:\:\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$

Question Number 15504 Answers: 1 Comments: 3

Solve ⌈x^2 ⌉=(⌊x⌋)^2 +2x

$$\mathrm{Solve}\:\lceil{x}^{\mathrm{2}} \rceil=\left(\lfloor{x}\rfloor\right)^{\mathrm{2}} +\mathrm{2}{x} \\ $$

Question Number 15501 Answers: 1 Comments: 0

Solve x^3 −⌊x⌋=3

$$\mathrm{Solve} \\ $$$${x}^{\mathrm{3}} −\lfloor{x}\rfloor=\mathrm{3} \\ $$

Question Number 15497 Answers: 1 Comments: 5

P=Σ_(n∈P) ^∞ n Q=Σ_(n∉P) ^∞ n P=2+3+5+7+... Q=1+4+6+8+... Is P>Q? Is Q>P?

$${P}=\underset{{n}\in\mathbb{P}} {\overset{\infty} {\sum}}{n}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}=\underset{{n}\notin\mathbb{P}} {\overset{\infty} {\sum}}{n} \\ $$$${P}=\mathrm{2}+\mathrm{3}+\mathrm{5}+\mathrm{7}+... \\ $$$${Q}=\mathrm{1}+\mathrm{4}+\mathrm{6}+\mathrm{8}+... \\ $$$$\: \\ $$$$\mathrm{Is}\:{P}>{Q}?\:\:\:\mathrm{Is}\:{Q}>{P}? \\ $$

Question Number 15234 Answers: 0 Comments: 2

A question related to Q.15184 Find the maximum of f(x)=(ln x)^(1/x)

$$\mathrm{A}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{Q}.\mathrm{15184} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{ln}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$

Question Number 15051 Answers: 1 Comments: 0

Solve simultaneously x + y + z = 6 ............ equation (i) x^3 + y^3 + z^3 = 92 .......... equation (ii) x − y = z ........... equation (iii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{92}\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}\:−\:\mathrm{y}\:=\:\mathrm{z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\:\mathrm{equation}\:\left(\mathrm{iii}\right) \\ $$

Question Number 14988 Answers: 0 Comments: 2

Solve on Z_4 ax+b=[0]_4 a,b∈Z_4 ax^2 +bx+c=[0]_4 a,b,c∈Z_4

$${Solve}\:{on}\:\mathbb{Z}_{\mathrm{4}} \: \\ $$$${ax}+{b}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b}\in\mathbb{Z}_{\mathrm{4}} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b},{c}\in\mathbb{Z}_{\mathrm{4}} \\ $$

Question Number 14811 Answers: 1 Comments: 0

why (√x^2 )=∣x∣ ?

$${why}\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid\:\:\:\:?\: \\ $$

Question Number 14775 Answers: 1 Comments: 0

Using the remainder theorem to factorize completely the expression x^3 (y − z) + y^3 (z − x) + z^3 (x − y)

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{factorize}\:\mathrm{completely}\:\mathrm{the}\:\mathrm{expression}\: \\ $$$$\mathrm{x}^{\mathrm{3}} \left(\mathrm{y}\:−\:\mathrm{z}\right)\:+\:\mathrm{y}^{\mathrm{3}} \left(\mathrm{z}\:−\:\mathrm{x}\right)\:+\:\mathrm{z}^{\mathrm{3}} \left(\mathrm{x}\:−\:\mathrm{y}\right)\: \\ $$

Question Number 14747 Answers: 1 Comments: 0

ε>0 6−ε≤xy≤6+ε 5−ε≤x+y≤5+ε Find x & y

$$\epsilon>\mathrm{0} \\ $$$$\mathrm{6}−\epsilon\leqslant{xy}\leqslant\mathrm{6}+\epsilon \\ $$$$\mathrm{5}−\epsilon\leqslant{x}+{y}\leqslant\mathrm{5}+\epsilon \\ $$$${Find}\:{x}\:\&\:{y} \\ $$

Question Number 14594 Answers: 1 Comments: 0

Question Number 14559 Answers: 1 Comments: 0

Solve for x ((6x + 2a + 3b + c )/(6x + 2a − 3b − c)) = ((2x + 6a + b + 3c)/(2x + 6a − b − 3c))

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x} \\ $$$$\frac{\mathrm{6x}\:+\:\mathrm{2a}\:+\:\mathrm{3b}\:+\:\mathrm{c}\:}{\mathrm{6x}\:+\:\mathrm{2a}\:−\:\mathrm{3b}\:−\:\mathrm{c}}\:=\:\frac{\mathrm{2x}\:+\:\mathrm{6a}\:+\:\mathrm{b}\:+\:\mathrm{3c}}{\mathrm{2x}\:+\:\mathrm{6a}\:−\:\mathrm{b}\:−\:\mathrm{3c}} \\ $$

Question Number 14535 Answers: 2 Comments: 6

Question Number 14521 Answers: 0 Comments: 0

Question Number 14491 Answers: 1 Comments: 0

(√(25))

$$\sqrt{\mathrm{25}} \\ $$$$ \\ $$

Question Number 14483 Answers: 0 Comments: 8

x^y +y^x =3.....(1) x+y=3.....(2) solve the equation

$$\mathrm{x}^{\mathrm{y}} +\mathrm{y}^{\mathrm{x}} =\mathrm{3}.....\left(\mathrm{1}\right) \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{3}.....\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$

Question Number 14435 Answers: 0 Comments: 0

∫e^(−x^2 ) dx=?

$$\int{e}^{−{x}^{\mathrm{2}} } {dx}=? \\ $$

Question Number 14398 Answers: 1 Comments: 0

Solve: (7/2) + ((3y)/(x + y)) = (√x) + 4(√y) .......... equation (i) (x^2 + y^2 )(x + 1) = 4 + 2xy(x − 1) .......... equation (ii)

$$\mathrm{Solve}:\: \\ $$$$\frac{\mathrm{7}}{\mathrm{2}}\:+\:\frac{\mathrm{3y}}{\mathrm{x}\:+\:\mathrm{y}}\:=\:\sqrt{\mathrm{x}}\:+\:\mathrm{4}\sqrt{\mathrm{y}}\:\:\:\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{4}\:+\:\mathrm{2xy}\left(\mathrm{x}\:−\:\mathrm{1}\right)\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Question Number 14396 Answers: 1 Comments: 2

Question Number 14354 Answers: 1 Comments: 0

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