Решите уравнение 20y/36y^2-4 - 2y-3/2-6y=5-2y/6y+2. Created by Naryy. algebra-ru

6y+2...

Naryy @Naryy

July 2022 1 14 Report

Решите уравнение
20y/36y^2-4 - 2y-3/2-6y=5-2y/6y+2

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Ответ:

{1/6; 1}

Объяснение:

$\tt \displaystyle \frac{20 \cdot y}{36 \cdot y^2-4}-\frac{2 \cdot y-3}{2-6 \cdot y} =\frac{5-2 \cdot y}{6 \cdot y+2}$

Область допустимых значений:

$\tt \displaystyle \left \{\begin{array}{ccc}36 \cdot y^2-4 \neq 0\\2-6 \cdot y\neq 0\\6 \cdot y+2\neq 0\end{array} \Leftrightarrow \left \{\begin{array}{ccc}(6 \cdot y-2) \cdot (6 \cdot y+2) \neq 0\\6 \cdot y\neq 2\\6 \cdot y\neq -2\end{array} \Leftrightarrow \left \{ {{y\neq \dfrac{1}{3} } \atop {y\neq -\dfrac{1}{3}}} \right.$

$\tt \displaystyle \frac{20 \cdot y}{(6 \cdot y)^2-2^2}+\frac{2 \cdot y-3}{6 \cdot y-2} +\frac{2 \cdot y-5}{6 \cdot y+2}=0$

$\tt \displaystyle \frac{20 \cdot y}{(6 \cdot y-2) \cdot (6 \cdot y+2)}+\frac{2 \cdot y-3}{6 \cdot y-2} +\frac{2 \cdot y-5}{6 \cdot y+2}=0 \;\;\; | \cdot (6 \cdot y-2)\cdot (6 \cdot y+2)$

20·y+(2·y-3)·(6·y+2)+(2·y-5)·(6·y-2)=0

20·y+12·y²+4·y-18·y-6+12·y²-4·y-30·y+10=0

24·y²-28·y+4=0 |:4

6·y²-7·y+1=0:

D=(-7)²-4·6·1=49-24=25=5²,

y₁=(7-5)/(2·6)=2/12=1/6 - подходит,

y₂=(7+5)/(2·6)=12/12=1 - подходит.

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