№ 4 ЗПС Алгебра = № 4 ЗПС Математика
Доведіть тотожність:
1. $\frac{\left(x^2-\frac{1}{y^2}\right)^x\cdot\left(y+\frac{1}{x}\right)^{y-x}}{\left(y^2-\frac{1}{x^2}\right)^y\cdot\left(x-\frac{1}{y}\right)^{x-y}}=\left(\frac{x}{y}\right)^{x+y};$
2. $\left(1+\frac{b^2+c^2-a^2}{2bc}\right):$
$: \left(\left(\frac{1}{a}+\frac{1}{b+c}\right)\right.:$
$: \left.\left(\frac{1}{a}-\frac{1}{b+c}\right)\right)=\frac{(b+c-a)^2}{2bc};$
3. $\frac{(x-y)^2+xy}{(x-y)^2-xy}:$
$: \frac{x^5+y^5+x^2y^3+x^3y^2}{\left(x^3+y^3+x^2y+xy^2\right)\left(x^3-y^3\right)}=$
$= x-y;$
4. $\left(\frac{2-y}{y-1}+\frac{2(x-1)}{x-2}\right):$
$: \left(\frac{y(x-1)}{y-1}+\frac{x(2-y)}{x-2}\right)=\frac{1}{x-y}.$
Розв'язок:
1. $\frac{\left(x^2-\frac{1}{y^2}\right)^x\cdot\left(y+\frac{1}{x}\right)^{y-x}}{\left(y^2-\frac{1}{x^2}\right)^y\cdot\left(x-\frac{1}{y}\right)^{x-y}}=\left(\frac{x}{y}\right)^{x+y}.$
а) $\left(x^2-\frac{1}{y^2}\right)^x=\left(\frac{x^2y^2-1}{y^2}\right)^x;$
б) $\left(y+\frac{1}{x}\right)^{y-x}=\left(\frac{xy+1}{x}\right)^{y-x}=$
$= \left(\frac{xy+1}{x}\right)^y:\left(\frac{xy+1}{x}\right)^x=$
$= \frac{(xy+1)^y\cdot x^x}{x^y\cdot(xy+1)^x};$
в) $\left(\frac{x^2y^2-1}{y^2}\right)^x\cdot\frac{(xy+1)^4\cdot x^x}{x^4\cdot(xy+1)^x}=$
$= \frac{(xy-1)^x\cdot(xy+1)^x\cdot(xy+1)^y\cdot x^x}{y^{2x}\cdot x^y\cdot(xy+1)^x}=$
$= \frac{(xy-1)^x(xy+1)^y\cdot x^x}{y^{2x}\cdot x^y};$
г) $\left(y^2-\frac{1}{x^2}\right)^y=\left(\frac{x^2y^2-1}{x^2}\right)^y;$
д) $\left(x-\frac{1}{y}\right)^{x-y}=\left(\frac{xy-1}{y}\right)^{x-y}=$
$= \frac{(xy-1)^x}{y^x}:\frac{(xy-1)^y}{y^y}=$
$= \frac{(xy-1)^x\cdot y^y}{y^x\cdot(xy-1)^y};$
е) $\left(y^2-\frac{1}{x^2}\right)^y\cdot\left(x-\frac{1}{y}\right)^{x-y}=$
$= \frac{(xy-1)^y(xy+1)^y\cdot(xy-1)^x\cdot y^y}{x^{2y}\cdot y^x(xy-1)^y}=$
$= \frac{(xy+1)^y(xy-1)^x\cdot y^y}{x^{2y}\cdot y^x};$
є) $\ \frac{(xy-1)^x(xy+1)^y\cdot x^x}{y^{2x}\cdot x^y}:$
$: \frac{(xy+1)^y(xy-1)^x\cdot y^y}{x^{2y}\cdot y^x}=$
$= \frac{(xy-1)^x\cdot(xy+1)^y\cdot x^x\cdot x^{2y}\cdot y^x}{y^{2x}\cdot x^y\cdot(xy+1)^y(xy-1)^x\cdot y^y}=$
$= \frac{x^x\cdot x^{2y}\cdot x^{-y}}{y^{2x}\cdot y^y\cdot y^{-x}}=$
$= \frac{x^{x+y}}{y^{x+y}}=\left(\frac{x}{y}\right)^{x+y};$
2. $\left(1+\frac{b^2+c^2-a^2}{2bc}\right):$
$: \left(\left(\frac{1}{a}+\frac{1}{b+c}\right)\right.:$
$: \left.\left(\frac{1}{a}-\frac{1}{b+c}\right)\right)=\frac{(b+c-a)^2}{2bc};$
а) $\ 1+\frac{b^2+c^2-a^2}{2bc}=\frac{\left(2bc+b^2+c^2\right)-a^2}{2bc}=$
$= \frac{(b+c)^2-a^2}{2ba}=\frac{(b+c-a)(b+c+a)}{2bc};$
б) $\frac{1}{a}+\frac{1}{b+c}=\frac{b+c+a}{a(b+c)};$
в) $\frac{1}{a}-\frac{1}{b+c}=\frac{b+c-a}{a(b+c)};$
г) $\frac{b+c+a}{a(b+c)}:\frac{b+c-a}{a(b+c)}=$
$= \frac{(a+b+c)\cdot a(b+c)}{a(b+c)\cdot(b+c-a)}= \frac{a+b+c}{b+c-a};$
д) $\frac{(b+c-a)(b+c+a)}{2bc}:\frac{a+b+c}{b+c-a}=$
$= \frac{(b+c-a)(b+c+a)\cdot(b+c-a)}{2bc\cdot(a+b+c)}=$
$= \frac{(b+c-a)^2}{2bc};$
3. $\frac{(x-y)^2+xy}{(x-y)^2-xy}:$
$: \frac{x^5+y^5+x^2y^3+x^3y^2}{\left(x^3+y^3+x^2y+xy^2\right)\left(x^3-y^3\right)}=$
$= x-y;$
$\frac{x^2+y^2-xy}{x^2+y^2+xy}\cdot$
$\cdot\frac{\left(x^3+y^3+xy(x+y)\right)\left(x^3-y^3\right.}{\left(x^5+x^3y^3\right)+\left(\left(y^5+x^2y^3\right)\right.}=$
$= \frac{x^2+y^2-xy}{x^2+y^2+xy}\cdot$
$\cdot\frac{(x+y)\left(x^2-xy+y^2+xy\right)\left(x^3-y^3\right)}{x^3\left(x^2+y^2\right)+y^3\left(y^2+x^2\right)}=$
$= \frac{x^2-xy+y^2}{x^2+xy+y^2}\cdot$
$\cdot\frac{(x+y)\left(x^2+y^2\right)(x-y)\left(x^2+xy+y^2\right)}{\left(x^2+y^2\right)\left(x^3+y^3\right)}=$
$= \frac{\left(x^3+y^3\right)(x-y)}{x^3+y^3}=x-y;$
4. $\left(\frac{2-y}{y-1}+\frac{2(x-1)}{x-2}\right):$
$: \left(\frac{y(x-1)}{y-1}+\frac{x(2-y)}{x-2}\right)=\frac{1}{x-y}.$
а) $\frac{2-y}{y-1}+\frac{2(x-1)}{x-2}=$
$= \frac{(2-y)(x-2)+2(x-1)(y-1)}{(y-1)(x-2)}=$
$= \frac{2x-4-xy+2y+2xy-2x-2y+2}{(y-1)(x-2)}=$
$= \frac{xy-2}{(y-1)(x-2)};$
б) $\frac{y(x-1)}{y-1}+\frac{x(2-y)}{x-2}=$
$= \frac{y(x-1)(x-2)+x(2-y)(y-1)}{(y-1)(x-2)}=$
$= \frac{2y-2x+x^2y-y^2x}{(y-1)(x-2)}=$
$= \frac{2(y-x)+xy(x-y)}{(y-1)(x-2)}=$
$= \frac{2(y-x)-xy(y-x)}{(y-1)(x-2)}=$
$= \frac{(y-x)(2-xy)}{(y-1)(x-2)};$
в) $\frac{xy-2}{(y-1)(x-2)}:\frac{(y-x)(2-xy)}{(y-1)(x-2)}=$
$= -\frac{2-xy}{(y-1)(x-2)}\cdot\frac{(y-1)(x-2)}{(y-x)(2-xy)}=$
$= -\frac{1}{y-x}=\frac{1}{x-y}.$