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AlgebraQuestion and Answers: Page 235

Question Number 105822    Answers: 2   Comments: 0

Given f(x+(1/(2x))) = x^2 +(1/(4x^2 ))+3x+(3/(2x)) find f(2)

$$\mathcal{G}{iven}\:{f}\left({x}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)\:=\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }+\mathrm{3}{x}+\frac{\mathrm{3}}{\mathrm{2}{x}} \\ $$$${find}\:{f}\left(\mathrm{2}\right)\: \\ $$

Question Number 105704    Answers: 1   Comments: 2

Solve the following system of equations: { ((x^3 y+x^3 y^2 +2x^2 y^2 +x^2 y^3 +xy^3 =30)),((x^2 y+xy+x+y+xy^2 =11)) :}

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \mathrm{y}+\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} +\mathrm{xy}^{\mathrm{3}} =\mathrm{30}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{xy}+\mathrm{x}+\mathrm{y}+\mathrm{xy}^{\mathrm{2}} =\mathrm{11}}\end{cases} \\ $$

Question Number 105687    Answers: 3   Comments: 0

{ ((x^2 +y^2 = 13)),((x^3 +y^3 = 35 )) :}

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{13}}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} \:=\:\mathrm{35}\:}\end{cases} \\ $$

Question Number 105686    Answers: 1   Comments: 0

solve 7^x + 24^x = 25^x

$${solve}\:\mathrm{7}^{{x}} \:+\:\mathrm{24}^{{x}} \:=\:\mathrm{25}^{{x}} \: \\ $$

Question Number 105634    Answers: 0   Comments: 0

Question Number 105625    Answers: 1   Comments: 0

prove that ((m),(m) ) ((( n)),((n−k)) )+ ((( m)),((m−1)) ) ((( n)),((n−k+1)) ) + ((( m)),((m−2)) ) ((( n)),((n−k+2)) )+......+ ((( m)),((m−k)) ) ((n),(n) ) = (((m+n)),(( k)) )

$${prove}\:{that} \\ $$$$\begin{pmatrix}{{m}}\\{{m}}\end{pmatrix}\begin{pmatrix}{\:\:\:{n}}\\{{n}−{k}}\end{pmatrix}+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{1}}\end{pmatrix} \\ $$$$+\begin{pmatrix}{\:\:\:{m}}\\{{m}−\mathrm{2}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{2}}\end{pmatrix}+......+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{m}+{n}}\\{\:\:\:\:\:{k}}\end{pmatrix} \\ $$

Question Number 105605    Answers: 2   Comments: 0

prove by mathematical induction 2^3 +4^3 +6^3 +8^3 +...+(2n)^3 = 2n^2 (n+1)^2

$${prove}\:{by}\:{mathematical}\:{induction}\: \\ $$$$\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +\mathrm{8}^{\mathrm{3}} +...+\left(\mathrm{2}{n}\right)^{\mathrm{3}} =\:\mathrm{2}{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Question Number 105536    Answers: 1   Comments: 0

Question Number 105534    Answers: 1   Comments: 2

Question Number 105456    Answers: 2   Comments: 0

Prove that the curve y=x^4 +3x^2 +2x does not meet the straight line y=2x−1 and find the distace between their nearest points.(Answer (1/(√5)))

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{4}} +\mathrm{3x}^{\mathrm{2}} +\mathrm{2x} \\ $$$$\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{y}=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distace}\:\mathrm{between} \\ $$$$\mathrm{their}\:\mathrm{nearest}\:\mathrm{points}.\left(\mathrm{Answer}\:\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right) \\ $$

Question Number 105448    Answers: 3   Comments: 2

Given { ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z)),((((4z^2 )/(4z^2 +1)) = x)) :} . find x+y+z ?

$$\mathcal{G}{iven}\:\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases}\:\:\:.\:{find} \\ $$$${x}+{y}+{z}\:?\: \\ $$

Question Number 105414    Answers: 0   Comments: 2

Question Number 105413    Answers: 0   Comments: 0

Question Number 105406    Answers: 0   Comments: 0

Question Number 105340    Answers: 1   Comments: 0

1\Show that the polynomial, P(x)=x^n +ax+b, n≥2 (a,b)∈R^2 has at most 3 real distinct roots. 2\Show that the equation x^4 +4ax+b=0 has no more than 2 real distinct roots.

$$\mathrm{1}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polynomial},\:\mathrm{P}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} +\mathrm{ax}+\mathrm{b},\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \:\mathrm{has}\:\mathrm{at}\:\mathrm{most}\:\mathrm{3}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$$$\mathrm{2}\backslash\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{4ax}+\mathrm{b}=\mathrm{0}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{more}\:\mathrm{than}\:\mathrm{2}\:\mathrm{real}\:\mathrm{distinct}\:\mathrm{roots}. \\ $$

Question Number 105322    Answers: 0   Comments: 1

Question Number 105320    Answers: 0   Comments: 0

find C kx^2 +ky^2 +z^2 ≥C x,y,z,k∈R such that xy+yz+zx=1

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{C}} \\ $$$${kx}^{\mathrm{2}} +{ky}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{C} \\ $$$${x},{y},{z},{k}\in\mathbb{R}\:\:{such}\:{that} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$

Question Number 105234    Answers: 0   Comments: 0

f:R→R x,y∈R f(f(x)f(y))+f(x+y)=f(xy) f(x)=?

$${f}:\mathbb{R}\rightarrow\mathbb{R}\:\:{x},{y}\in\mathbb{R} \\ $$$${f}\left({f}\left({x}\right){f}\left({y}\right)\right)+{f}\left({x}+{y}\right)={f}\left({xy}\right) \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 105222    Answers: 0   Comments: 0

Question Number 105189    Answers: 0   Comments: 2

Question Number 105158    Answers: 4   Comments: 0

{ ((xy+x+y = 20)),((x^2 +y^2 = 40)) :}find x and y

$$\begin{cases}{{xy}+{x}+{y}\:=\:\mathrm{20}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\mathrm{40}}\end{cases}{find}\:{x}\:{and}\:{y} \\ $$

Question Number 105124    Answers: 1   Comments: 0

x^4 +ax^3 +bx^2 +cx+d=0 Find x.

$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${Find}\:{x}. \\ $$

Question Number 105120    Answers: 1   Comments: 0

f(2f(x)+f(y))=2x+y f:R→R f(x)=? x,y∈R

$${f}\left(\mathrm{2}{f}\left({x}\right)+{f}\left({y}\right)\right)=\mathrm{2}{x}+{y}\:\:\:{f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=?\:\:\:\:{x},{y}\in\mathbb{R} \\ $$$$ \\ $$

Question Number 105083    Answers: 0   Comments: 0

Question Number 105082    Answers: 0   Comments: 0

Question Number 105023    Answers: 1   Comments: 0

(((x^2 −y^2 )^2 )/((1−x^2 )(y^2 −1)))+(((1−x^2 )^2 )/((x^2 −y^2 )(y^2 −1)))+(((y^2 −1)^2 )/((1−x^2 )(x^2 −y^2 )))=

$$\frac{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}= \\ $$$$ \\ $$

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