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Goshia [24]
2 years ago
8

AHHHHHHHHHHHH this is so hard

Mathematics
2 answers:
Llana [10]2 years ago
8 0

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \cfrac{ {a}^{2} - 64 }{ {a}^{2}  - 10a + 24}  \sdot \cfrac{ {a}^{2} - 12a + 36 }{ {a}^{2}  + 4a - 32}

\qquad \sf  \dashrightarrow \:  \cfrac{( {a}^{}  + 8)(a - 8) }{ {a}^{2}  - 6a  - 4a+ 24}  \sdot \cfrac{ {a}^{2} - 6a - 6a + 36 }{ {a}^{2}  + 8a - 4a- 32}

\qquad \sf  \dashrightarrow \:  \cfrac{( {a}^{}  + 8)(a - 8) }{ {a}^{} (a - 6)  - 4(a - 6)}  \sdot \cfrac{ {a}^{} (a - 6) - 6(a  - 6) }{ {a}^{}(a  +  8) - 4(a + 8)}

\qquad \sf  \dashrightarrow \:  \cfrac{( {a}^{}  + 8)(a - 8) }{(a - 6) (a -4)}  \sdot \cfrac{ (a - 6) (a  - 6) }{ {}^{}(a   - 4)(a + 8)}

\qquad \sf  \dashrightarrow \:  \cfrac{(a - 8) }{(a -4)}  \sdot \cfrac{ (a  - 6) }{ {}^{}(a   - 4)}

\qquad \sf  \dashrightarrow \:  \cfrac{(a - 8)(a - 6) }{(a -4) {}^{2} }

Or [ in expanded form ]

\qquad \sf  \dashrightarrow \:  \cfrac{ {a}^{2} - 8a - 6a + 48 }{ {a}^{2} - 8a + 16 }

\qquad \sf  \dashrightarrow \:  \cfrac{ {a}^{2} -14a + 48 }{ {a}^{2} - 8a + 16 }

tatyana61 [14]2 years ago
7 0

Answer:

\dfrac{(a-8)(a-6)}{(a-4)^2}

Step-by-step explanation:

                           

<u>Given expression</u>:

\dfrac{a^2-64}{a^2-10a+24} \cdot \dfrac{a^2-12a+36}{a^2+4a-32}

<u>Factor the numerator and denominator of both fractions:</u>

\textsf{Apply the Difference of Two Squares formula} \:\:\:x^2-y^2=(x-y)(x+y):

\begin{aligned} a^2-64 & =a^2+8^2 \\ & =(a-8)(a+8)\end{aligned}

\begin{aligned}a^2-10a+24 & =a^2-4a-6a+24\\& = a(a-4)-6(a-4)\\ & = (a-6)(a-4) \end{aligned}

\begin{aligned}a^2-12a+36 & =a^2-6a-6a+36\\& = a(a-6)-6(a-6)\\ & = (a-6)(a-6) \end{aligned}

\begin{aligned}a^2+4a-32 & =a^2+8a-4a-32\\& = a(a+8)-4(a+8)\\ & = (a-4)(a+8) \end{aligned}

Therefore:

\dfrac{(a-8)(a+8)}{(a-6)(a-4)} \cdot \dfrac{(a-6)(a-6)}{(a-4)(a+8)}

\textsf{Apply the fraction rule}: \quad \dfrac{a}{b} \cdot \dfrac{c}{d}=\dfrac{ac}{bd}

\dfrac{(a-8)(a+8)(a-6)(a-6)}{(a-6)(a-4)(a-4)(a+8)}

Cancel the common factors (a + 8) and (a - 6):

\dfrac{(a-8)(a-6)}{(a-4)(a-4)}

Simplify the numerator:

\dfrac{(a-8)(a-6)}{(a-4)^2}

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