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AlgebraQuestion and Answers: Page 227

Question Number 112802 Answers: 3 Comments: 0

If xy+yz+zx=0, then ((1/(x^2 −yz))+(1/(y^2 −zx))+(1/(z^2 −xy)))(x,y,z ≠ 0) is equal to

$$\mathrm{If}\:\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=\mathrm{0},\:\mathrm{then} \\ $$$$ \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{yz}}+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} −\mathrm{zx}}+\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} −\mathrm{xy}}\right)\left(\mathrm{x},\mathrm{y},\mathrm{z}\:\neq\:\mathrm{0}\right)\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to} \\ $$$$ \\ $$

Question Number 112796 Answers: 0 Comments: 2

For any real numbers x,y and z if x+y+z=2, then prove that xyz≥8(1−x)(1−y)(1−z).

$${For}\:{any}\:{real}\:{numbers}\:{x},{y}\:{and}\:{z}\:{if} \\ $$$${x}+{y}+{z}=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$${xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right). \\ $$

Question Number 112795 Answers: 1 Comments: 1

Question Number 112626 Answers: 1 Comments: 0

Question Number 112625 Answers: 4 Comments: 1

If a+b=5, a^2 +b^2 =13, the value of a−b (where a>b) is

$$\mathrm{If}\:\mathrm{a}+\mathrm{b}=\mathrm{5},\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{13},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{a}−\mathrm{b}\:\left(\mathrm{where}\:\mathrm{a}>\mathrm{b}\right)\:\mathrm{is} \\ $$

Question Number 112624 Answers: 3 Comments: 0

Minimum value of x^2 +(1/(x^2 +1))−3 is

$$\mathrm{Minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{3}\:\mathrm{is} \\ $$

Question Number 112528 Answers: 0 Comments: 2

If ((n−7)/(n−17)) and ((n−14)/(n−24)) are integers numbers n=?

$${If}\:\:\:\:\:\frac{{n}−\mathrm{7}}{{n}−\mathrm{17}}\:\:\:\:\:\:\:{and}\:\:\:\:\frac{{n}−\mathrm{14}}{{n}−\mathrm{24}}\:\:\:{are}\:{integers}\:{numbers} \\ $$$${n}=? \\ $$

Question Number 112509 Answers: 0 Comments: 1

Question Number 112478 Answers: 1 Comments: 0

If P(x)= x^3 +ax^2 +bx+c, with a, b and c real numbers. Roots of P(x) z+3i, z+9i and 2z−4, find ∣a+b+c∣. Note: z is complex number.

$${If}\:{P}\left({x}\right)=\:{x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c},\:{with}\:{a},\:{b}\:{and}\:{c}\:{real}\:{numbers}. \\ $$$${Roots}\:{of}\:{P}\left({x}\right)\:{z}+\mathrm{3}{i},\:{z}+\mathrm{9}{i}\:{and}\:\mathrm{2}{z}−\mathrm{4},\:{find}\:\mid{a}+{b}+{c}\mid. \\ $$$$\boldsymbol{{N}}{ote}:\:{z}\:{is}\:{complex}\:{number}. \\ $$

Question Number 112464 Answers: 1 Comments: 0

Question Number 112461 Answers: 1 Comments: 0

If x, y and z are numbers integers different of 0 { ((49x+7y+z=0)),((25x−5y+z=0)) :} find (√(y^2 −4xz))

$${If}\:\:\:{x},\:{y}\:{and}\:{z}\:\:{are}\:{numbers}\:\:{integers}\:{different}\:{of}\:\:\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{49}{x}+\mathrm{7}{y}+{z}=\mathrm{0}}\\{\mathrm{25}{x}−\mathrm{5}{y}+{z}=\mathrm{0}}\end{cases}\:\:\:\:\:\:\:\:\:{find}\:\sqrt{{y}^{\mathrm{2}} −\mathrm{4}{xz}} \\ $$

Question Number 112408 Answers: 0 Comments: 1

Question Number 112360 Answers: 1 Comments: 1

(((x^3 +x)/3))^3 +(((x^3 +x)/3))= 3x

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

Question Number 112280 Answers: 2 Comments: 0

If 15 ≤ ∣10−(1/3)a∣ < 20 find a ∈ R

$$\:\:\mathrm{If}\:\mathrm{15}\:\leqslant\:\mid\mathrm{10}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{a}\mid\:<\:\mathrm{20}\: \\ $$$$\mathrm{find}\:\mathrm{a}\:\in\:\mathbb{R} \\ $$

Question Number 112097 Answers: 0 Comments: 2

∣∣x−1∣+1∣ < x

$$\:\:\:\:\:\:\mid\mid{x}−\mathrm{1}\mid+\mathrm{1}\mid\:<\:{x} \\ $$

Question Number 112087 Answers: 2 Comments: 0

a(√a) −3 = 10(√a) → (√a) + (√a^(−1) ) =?

$$\:\:\:\mathrm{a}\sqrt{\mathrm{a}}\:−\mathrm{3}\:=\:\mathrm{10}\sqrt{\mathrm{a}}\:\rightarrow\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{a}^{−\mathrm{1}} }\:=? \\ $$

Question Number 112065 Answers: 1 Comments: 0

Question Number 112058 Answers: 1 Comments: 0

Question Number 112053 Answers: 1 Comments: 1

solve for x, y, z ∈C: 2x^2 −3x=(√(13x^2 −52x+40)) 6y^2 −14x=(√(x^2 −220x+300)) z^2 −2z=(√(−12x^2 +72x−132)) [exact solutions possible in all cases]

$$\mathrm{solve}\:\mathrm{for}\:{x},\:{y},\:{z}\:\in\mathbb{C}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{3}{x}=\sqrt{\mathrm{13}{x}^{\mathrm{2}} −\mathrm{52}{x}+\mathrm{40}} \\ $$$$\mathrm{6}{y}^{\mathrm{2}} −\mathrm{14}{x}=\sqrt{{x}^{\mathrm{2}} −\mathrm{220}{x}+\mathrm{300}} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}=\sqrt{−\mathrm{12}{x}^{\mathrm{2}} +\mathrm{72}{x}−\mathrm{132}} \\ $$$$\left[\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}\:\mathrm{in}\:\mathrm{all}\:\mathrm{cases}\right] \\ $$

Question Number 112008 Answers: 0 Comments: 0

Question Number 111939 Answers: 2 Comments: 0

Question Number 111982 Answers: 2 Comments: 0

Question Number 111909 Answers: 2 Comments: 0

solve { ((x(√x)+y(√y) = 5)),((x(√y) +y(√x) = 1)) :}

$${solve}\:\begin{cases}{{x}\sqrt{{x}}+{y}\sqrt{{y}}\:=\:\mathrm{5}}\\{{x}\sqrt{{y}}\:+{y}\sqrt{{x}}\:=\:\mathrm{1}}\end{cases} \\ $$

Question Number 111724 Answers: 0 Comments: 6

How many real numbers x satisfy the equation 3^(2x+2) −3^(x+3) −3^x +3=0 ?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{x}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{3}^{\mathrm{2x}+\mathrm{2}} −\mathrm{3}^{\mathrm{x}+\mathrm{3}} −\mathrm{3}^{\mathrm{x}} +\mathrm{3}=\mathrm{0}\:? \\ $$

Question Number 111623 Answers: 1 Comments: 2

(x−2)(x+3)(x−1)^2 ≥ 0

$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{0} \\ $$

Question Number 111601 Answers: 1 Comments: 0

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