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Category: Algebra

Inequality-relation-starting-a-new-thread-x-p-p-p-1-1-p-x-q-q-q-1-1-q-p-2-q-1-x-1-x-p-p-p-1-1-6-x-q-q-q-1-1-2-x-p-p-p-1-1-p-1-6-1-2-1-3-x-p-p-p-

Question Number 1952 by prakash jain last updated on 25/Oct/15 $$\mathrm{Inequality}\:\mathrm{relation}\:\mathrm{starting}\:\mathrm{a}\:\mathrm{new}\:\mathrm{thread} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\geqslant\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{q}} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{1},\:{x}=\mathrm{1} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\frac{{x}^{{p}}…

Is-necessary-and-suficient-for-two-inequalities-to-be-equivalent-If-a-gt-b-Are-A-gt-B-and-A-a-gt-B-b-equivalent-

Question Number 1937 by Rasheed Soomro last updated on 25/Oct/15 $$\bullet{Is}\:\:\:'\Leftrightarrow'\:\:{necessary}\:{and}\:{suficient}\:{for}\:{two} \\ $$$${inequalities}\:{to}\:{be}\:{equivalent}? \\ $$$$\bullet{If}\:\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:: \\ $$$${Are}\:\:\boldsymbol{\mathrm{A}}>\boldsymbol{\mathrm{B}}\:{and}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{a}}\:>\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{b}}\:{equivalent}? \\ $$ Answered by 123456 last updated on…

f-2-x-f-x-2-2-f-2-x-stands-for-f-x-2-f-x-If-possible-solve-stepwise-

Question Number 1902 by Rasheed Soomro last updated on 23/Oct/15 $${f}^{\:\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2}\:,\:{f}^{\:\mathrm{2}} \left({x}\right)\:{stands}\:{for}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({If}\:{possible}\:{solve}\:{stepwise}\right) \\ $$ Commented by 123456 last updated…

Consider-the-system-of-equations-2yz-zx-5xy-2-yz-zx-2xy-1-yz-2zx-6xy-3-Show-that-xyz-6-and-find-the-possible-values-of-x-y-and-z-

Question Number 1899 by Yozzy last updated on 22/Oct/15 $${Consider}\:{the}\:{system}\:{of}\:{equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{yz}+{zx}−\mathrm{5}{xy}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−{zx}+\mathrm{2}{xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−\mathrm{2}{zx}+\mathrm{6}{xy}=\mathrm{3}. \\ $$$${Show}\:{that}\:{xyz}=\pm\mathrm{6}\: \\ $$$${and}\:{find}\:{the}\:{possible}\:{values} \\ $$$${of}\:{x},{y}\:{and}\:{z}. \\ $$ Commented…

Question-132961

Question Number 132961 by danielasebhofoh last updated on 17/Feb/21 Answered by Ar Brandon last updated on 17/Feb/21 $$\mathrm{2n}−\mathrm{3}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{3}} {\sum}}\mathrm{k}=\mathrm{2n}−\mathrm{3}+\frac{\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{4n}−\mathrm{6}+\mathrm{n}^{\mathrm{2}} −\mathrm{5n}+\mathrm{6}}{\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\overset{\mathrm{n}}…

Show-that-1-1-sinz-1-z-n-1-1-n-1-z-npi-1-z-npi-2-cotz-1-z-n-1-1-z-npi-1-z-npi-where-z-mpi-m-Z-given-the-fact-that-cosax-2sinapi-pi-1-2a-n-1-

Question Number 1838 by 112358 last updated on 11/Oct/15 $${Show}\:{that} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$\left(\mathrm{2}\right)\:{cotz}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$${where}\:{z}\neq{m}\pi,\:{m}\in\mathbb{Z}\:,\:{given}\:{the}\:{fact}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}}…