Question and Answers Forum

AlgebraQuestion and Answers: Page 2

Question Number 221944 Answers: 1 Comments: 1

Question Number 221943 Answers: 1 Comments: 0

Question Number 221863 Answers: 1 Comments: 1

If a and b are whole numbers such a^b =121 then find the value of (a−1)^(b+1)

$${If}\:{a}\:{and}\:{b}\:{are}\:{whole}\:{numbers}\:{such}\:{a}^{{b}} =\mathrm{121} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\left({a}−\mathrm{1}\right)^{{b}+\mathrm{1}} \\ $$

Question Number 221856 Answers: 0 Comments: 3

Question Number 221848 Answers: 0 Comments: 2

Question Number 221829 Answers: 4 Comments: 0

(√(70.71.72.73+1))

$$\sqrt{\mathrm{70}.\mathrm{71}.\mathrm{72}.\mathrm{73}+\mathrm{1}} \\ $$

Question Number 221782 Answers: 0 Comments: 0

(((81))^(1/((27))^(1/3^a ) ) )^((√4)) where a=4^0^4^3 and b=Σ_(n=1) ^6 n

$$\left(\sqrt[{\sqrt[{\mathrm{3}^{{a}} }]{\mathrm{27}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$$${where}\:{a}=\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } {and}\:{b}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{n} \\ $$

Question Number 221778 Answers: 0 Comments: 0

For ∀n∈N^∗ ,n≥3 Prove: ∫_0 ^1 (Σ_(2<p≤2n) e^(2πip+α) )^2 e^(−4πinα) dα>0,p is a prime number

$$\mathrm{For}\:\forall{n}\in\boldsymbol{{N}}^{\ast} ,{n}\geq\mathrm{3}\:\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{\mathrm{2}<{p}\leq\mathrm{2}{n}} {\sum}{e}^{\mathrm{2}\pi{ip}+\alpha} \right)^{\mathrm{2}} {e}^{−\mathrm{4}\pi{in}\alpha} {d}\alpha>\mathrm{0},{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number} \\ $$

Question Number 221754 Answers: 1 Comments: 0

((81))^(1/((64))^(1/((27))^(1/( 3^4^0^4^3 )) ) ) )^((√4))

$$\left.\sqrt[{\sqrt[{\sqrt[{\:\mathrm{3}^{\mathrm{4}^{\mathrm{0}^{\mathrm{4}^{\mathrm{3}} } } } }]{\mathrm{27}}}]{\mathrm{64}}}]{\mathrm{81}}\right)^{\sqrt{\mathrm{4}}} \\ $$

Question Number 221647 Answers: 0 Comments: 1

solve for x. x^1 + x^2 + x^3 = 4096

$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$

Question Number 221618 Answers: 3 Comments: 0

solve for x 2^x +4^x =8^x

$${solve}\:{for}\:{x} \\ $$$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$

Question Number 221669 Answers: 2 Comments: 3

Question Number 221585 Answers: 6 Comments: 1

solve for x ∈R (x^3 −6)^3 =x+6

$${solve}\:{for}\:{x}\:\in{R} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} ={x}+\mathrm{6} \\ $$

Question Number 221544 Answers: 1 Comments: 0

((x−1))^(1/(x+1)) = ((x+1))^(1/(x−1)) , x real

$$\:\:\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}\:=\:\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:,\:{x}\:{real}\: \\ $$

Question Number 221504 Answers: 1 Comments: 0

Find x if (x^4 /4^x )=(4^x /x^4 ) . x∈R

$${Find}\:{x}\:{if}\:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{{x}^{\mathrm{4}} }\:\:.\:\:\:{x}\in\mathbb{R} \\ $$

Question Number 221501 Answers: 3 Comments: 0

solve for x: (√(a−(√(a+x))))+(√(a+(√(a−x))))=2x it′s possible to solve for a but x seems impossible to me

$${solve}\:{for}\:\mathrm{x}: \\ $$$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}=\mathrm{2x} \\ $$$${it}'{s}\:{possible}\:{to}\:{solve}\:{for}\:\mathrm{a}\:{but}\:\mathrm{x}\:{seems} \\ $$$${impossible}\:{to}\:{me} \\ $$

Question Number 221430 Answers: 1 Comments: 0

Question Number 221399 Answers: 0 Comments: 0

let a,b ≥ 0 and a + b + ab = 3 show that; ((38)/(55)) ≤ (1/(a^2 + 2)) + (1/(b^2 +2)) + (1/(a^2 + b^2 + 1)) ≤ 1 , ((463)/(812)) ≤ (1/(a^3 + 2)) + (1/(b^3 + 2)) + (1/(a^3 + b^3 + 1)) ≤ 1 , ((193)/(308)) ≤ (1/(a^2 + 2)) + (1/(b^3 + 2)) + (1/(a^3 + b^2 + 1)) ≤1 , and ((463)/(812)) ≤ (1/(a^2 + 2)) + (1/(b^3 + 2)) + (1/(a^2 + b^3 + 1)) ≤ ((11)/(10))

$$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\:\geqslant\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{ab}}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}; \\ $$$$\:\:\frac{\mathrm{38}}{\mathrm{55}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{2}} \:+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:,\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:\:,\:\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{193}}{\mathrm{308}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:\:\leqslant\mathrm{1}\:\:,\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{and}} \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{1}}\:\leqslant\:\frac{\mathrm{11}}{\mathrm{10}}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 221393 Answers: 0 Comments: 0

a, b, c are complex number and ∣a∣ = ∣b∣=∣c∣= 1 and (a^2 /(bc))+(b^2 /(ac)) +(c^2 /(ab)) = −1 where ∣.∣ is modules function then ∣a+b+c∣ can be (A) 0 (B) 1 (C) (3/2) (D) 2

$$\:\:\:\:\:{a},\:{b},\:{c}\:{are}\:{complex}\:{number}\:{and}\: \\ $$$$\:\:\:\mid{a}\mid\:=\:\mid{b}\mid=\mid{c}\mid=\:\mathrm{1}\:{and}\:\:\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ac}}\:+\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:−\mathrm{1} \\ $$$$\:\:\:\:{where}\:\mid.\mid\:{is}\:\:{modules}\:{function} \\ $$$${then}\:\mid{a}+{b}+{c}\mid\:{can}\:{be}\: \\ $$$$\left({A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\left({C}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\mathrm{2} \\ $$

Question Number 221377 Answers: 2 Comments: 0

Find: Ω =lim_(n→∞) (2 ((10))^(1/n) − 1)^n = ?

$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\mathrm{10}}\:−\:\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \:=\:? \\ $$

Question Number 221359 Answers: 2 Comments: 0