មេរៀន៖ កន្សោមពីជគណិត - ថ្នាក់ទី៩ ភាគ ៦ : Algebraic Expressions - Grade 9 - parts 6

មេរៀន៖ កន្សោមពីជគណិត - ថ្នាក់ទី៩ (Algebraic Expressions - Grade 9)
(Full Step-by-Step Answer Key)

ភាគ ៦៖ ប្រភាគសនិទាន (Part 6: Rational Expressions)

រៀបរៀងដោយ៖ លោកគ្រូ CHEANG SOKKONG

💡 គន្លឹះ (Tip): នៅក្នុងភាគនេះ យើងនឹងសិក្សាពីការគណនា និងសម្រួល **ប្រភាគសនិទាន** ដោយត្រូវចេះតម្រូវភាគបែងរួម និងបំប្លែងភាគយក ឬភាគបែងជាផលគុណកត្តាសិនមុននឹងសម្រួល។
👉 បើកន្សោមវែងពេក ប្អូនៗអាចអូសទៅឆ្វេង-ស្ដាំបាន។

៣៩. ចូរកំណត់ a, b, c

ក. កំណត់ $a, b, c$ ដែល $\displaystyle \frac{2x^2-x+3}{x-1} = ax+b+\frac{c}{x-1}$

តម្រូវភាគបែងរួមអង្គទី២៖
$\displaystyle ax+b+\frac{c}{x-1} = \frac{(ax+b)(x-1)+c}{x-1}$
$\displaystyle = \frac{ax^2-ax+bx-b+c}{x-1}$
$\displaystyle = \frac{ax^2+(b-a)x+(c-b)}{x-1}$

ផ្ទឹមជាមួយអង្គទី១ $\displaystyle \frac{2x^2-x+3}{x-1}$ យើងបានប្រព័ន្ធសមីការ៖
$a = 2$
$b - a = -1 \Rightarrow b - 2 = -1 \Rightarrow b = 1$
$c - b = 3 \Rightarrow c - 1 = 3 \Rightarrow c = 4$

ដូចនេះ $a = 2, b = 1, c = 4$

៤០. ចូរគណនា និងសម្រួលកន្សោមខាងក្រោម

ក. $\displaystyle \frac{x-3}{3}-\frac{x+5}{4}$

$= \displaystyle \frac{4(x-3) - 3(x+5)}{12}$
$= \displaystyle \frac{4x - 12 - 3x - 15}{12}$
$= \displaystyle \frac{x - 27}{12}$

ខ. $\displaystyle 1+\frac{3x-4}{4}-\frac{x+2}{3}$

$= \displaystyle \frac{12 + 3(3x-4) - 4(x+2)}{12}$
$= \displaystyle \frac{12 + 9x - 12 - 4x - 8}{12}$
$= \displaystyle \frac{5x - 8}{12}$

គ. $\displaystyle \frac{x}{x+y}-\frac{x^2}{y^2-x^2}$

$= \displaystyle \frac{x}{x+y} - \frac{x^2}{(y-x)(y+x)}$
$= \displaystyle \frac{x(y-x) - x^2}{(y-x)(y+x)}$
$= \displaystyle \frac{xy - x^2 - x^2}{y^2-x^2}$
$= \displaystyle \frac{xy - 2x^2}{y^2-x^2}$

យ. $\displaystyle \frac{x^2-2x+1}{(x-1)^3}-\frac{x^2+x+1}{x^3-1}$

$= \displaystyle \frac{(x-1)^2}{(x-1)^3} - \frac{x^2+x+1}{(x-1)(x^2+x+1)}$
$= \displaystyle \frac{1}{x-1} - \frac{1}{x-1}$
$= 0$

៤១. ចូរគណនាកន្សោមសនិទានខាងក្រោម

ក. $\displaystyle \frac{6x-1}{81-x^2}-\frac{2x}{x+9}$

$= \displaystyle \frac{6x-1}{(9-x)(9+x)} - \frac{2x}{9+x}$
$= \displaystyle \frac{(6x-1) - 2x(9-x)}{(9-x)(9+x)}$
$= \displaystyle \frac{6x - 1 - 18x + 2x^2}{81-x^2}$
$= \displaystyle \frac{2x^2 - 12x - 1}{81-x^2}$

ខ. $\displaystyle \frac{10y-1}{100-y^2}-\frac{5y}{y+10}$

$= \displaystyle \frac{10y-1}{(10-y)(10+y)} - \frac{5y}{10+y}$
$= \displaystyle \frac{(10y-1) - 5y(10-y)}{(10-y)(10+y)}$
$= \displaystyle \frac{10y - 1 - 50y + 5y^2}{100-y^2}$
$= \displaystyle \frac{5y^2 - 40y - 1}{100-y^2}$

ឃ. $\displaystyle \frac{t}{2(t+3)}-\frac{2}{3(t+3)}$

$= \displaystyle \frac{3(t) - 2(2)}{6(t+3)}$
$= \displaystyle \frac{3t - 4}{6(t+3)}$

ង. $\displaystyle \frac{4x+16}{(x+4)(x-4)}+\frac{-4}{x+4}$

$= \displaystyle \frac{4(x+4)}{(x+4)(x-4)} - \frac{4}{x+4}$
$= \displaystyle \frac{4}{x-4} - \frac{4}{x+4}$
$= \displaystyle \frac{4(x+4) - 4(x-4)}{(x-4)(x+4)}$
$= \displaystyle \frac{4x + 16 - 4x + 16}{x^2-16}$
$= \displaystyle \frac{32}{x^2-16}$

៤២. ចូរសម្រួលកន្សោមសនិទានខាងក្រោម

ក. $\displaystyle \frac{\frac{1}{x}-\frac{3}{y}}{\frac{4}{y}}$

$= \displaystyle \frac{\frac{y-3x}{xy}}{\frac{4}{y}}$
$= \displaystyle \frac{y-3x}{xy} \times \frac{y}{4}$
$= \displaystyle \frac{y-3x}{4x}$

ខ. $\displaystyle \frac{\frac{4}{a^2}+\frac{b}{a}}{\frac{2}{a}-12}$

$= \displaystyle \frac{\frac{4+ab}{a^2}}{\frac{2-12a}{a}}$
$= \displaystyle \frac{4+ab}{a^2} \times \frac{a}{2(1-6a)}$
$= \displaystyle \frac{4+ab}{2a(1-6a)}$

គ. $\displaystyle \frac{\frac{2}{x}+\frac{3}{y}}{-\frac{5}{x}}$

$= \displaystyle \frac{\frac{2y+3x}{xy}}{-\frac{5}{x}}$
$= \displaystyle \frac{2y+3x}{xy} \times \left(-\frac{x}{5}\right)$
$= \displaystyle -\frac{2y+3x}{5y}$

ឈ. $\displaystyle \frac{\frac{8}{x}-9}{\frac{6}{y}}$

$= \displaystyle \frac{\frac{8-9x}{x}}{\frac{6}{y}}$
$= \displaystyle \frac{8-9x}{x} \times \frac{y}{6}$
$= \displaystyle \frac{y(8-9x)}{6x}$

ញ. $\displaystyle \frac{\frac{2b}{b-4}-\frac{3}{b^2}}{\frac{5}{5b-20}+\frac{3}{4b^2-16b}}$

ភាគយក៖ $\displaystyle \frac{2b}{b-4}-\frac{3}{b^2} = \frac{2b^3 - 3(b-4)}{b^2(b-4)} = \frac{2b^3-3b+12}{b^2(b-4)}$
ភាគបែង៖ $\displaystyle \frac{5}{5(b-4)} + \frac{3}{4b(b-4)} = \frac{1}{b-4} + \frac{3}{4b(b-4)} = \frac{4b+3}{4b(b-4)}$

ចែកប្រភាគ៖
$= \displaystyle \frac{\frac{2b^3-3b+12}{b^2(b-4)}}{\frac{4b+3}{4b(b-4)}}$
$= \displaystyle \frac{2b^3-3b+12}{b^2(b-4)} \times \frac{4b(b-4)}{4b+3}$
$= \displaystyle \frac{4(2b^3-3b+12)}{b(4b+3)}$

៤៣. ចូរគណនា និងសម្រួលកន្សោមខាងក្រោម

ក. $\displaystyle \frac{b^2+10b+21}{3(b^2-9)} \div \frac{2b^2+14b}{30b^2-90b}$

បំប្លែងកន្សោមនីមួយៗជាផលគុណកត្តា៖
$= \displaystyle \frac{(b+3)(b+7)}{3(b-3)(b+3)} \div \frac{2b(b+7)}{30b(b-3)}$
$= \displaystyle \frac{b+7}{3(b-3)} \times \frac{30b(b-3)}{2b(b+7)}$
សម្រួល $(b+7), (b-3)$ និង $b$ ចោល៖
$= \displaystyle \frac{30}{3 \times 2} = \frac{30}{6} = 5$

គ. $\displaystyle \frac{z^2+17z+66}{3(z^2-121)} \div \frac{2z^2+12z}{24z^2-264z}$

បំប្លែងកន្សោមនីមួយៗជាផលគុណកត្តា៖
$= \displaystyle \frac{(z+6)(z+11)}{3(z-11)(z+11)} \div \frac{2z(z+6)}{24z(z-11)}$
$= \displaystyle \frac{z+6}{3(z-11)} \times \frac{24z(z-11)}{2z(z+6)}$
សម្រួល $(z+6), (z-11)$ និង $z$ ចោល៖
$= \displaystyle \frac{24}{3 \times 2} = \frac{24}{6} = 4$