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bantu jawab plisss pake cara ya, makasii ​

Berikut ini adalah pertanyaan dari naswanfllh pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Bantu jawab plisss pake cara ya, makasii ​
bantu jawab plisss pake cara ya, makasii ​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jika a = 5, b = -¼, dan c = 2, maka hasil operasi:

  • {\sf{a.~ab^2c^5}=\textsf{\textbf{10}}}
  • {\sf{b.~10a^{-1}b^{-1}c^3=\boldsymbol-\textsf{\textbf{64}}}}
  • {\sf{c.~\dfrac{a^3b^{-2}c^2}{a^2b^{-1}c^4}=\boldsymbol-{\textsf{\textbf{5}}}}}
  • {\sf{d.~\dfrac{ab^{-3}c^{-4}+a^2bc^3}{abc^2}}={\textsf{\textbf{14}}}}

 

Pembahasan

Diketahui nilai a = 5, b = -¼, dan c = 2

 

\begin{aligned}{\sf{a.}}~&{\sf{ab^2c^5}}\\&={\sf{5\cdot\left(-\frac{1}{4}\right)^2\cdot2^5}}\\&={\sf{5\cdot\left(\frac{1}{2^2}\right)^2\cdot2^5}}\\&={\sf{5\cdot\left(2^{-2}\right)^2\cdot2^5}}\\&={\sf{5\cdot2^{-4}\cdot2^5}}\\&={\sf{5\cdot2^{-4+5}}}\\&={\sf{5\cdot2}}\\&={\sf{10}}\end{aligned}

 

\begin{aligned}{\sf{b.}}~&{\sf{10a^{-1}b^{-1}c^3}}\\&={\sf{10\cdot5^{-1}\cdot\left(-\frac{1}{4}\right)^{-1}\cdot2^3}}\\&={\sf{\cancel{10}\cdot\frac{1}{\cancel5}\cdot\left(-\frac{1}{2^2}\right)^{-1}\cdot2^3}}\\&={\sf{2\cdot\left(\left(-2\right)^{-2}\right)^{-1}\cdot2^3}}\\&={\sf{2\cdot\left(-2^2\right)\cdot2^3}}\\&={\sf{-\left(2\cdot2^2\cdot2^3\right)}}\\&={\sf{-\left(2^{1+2+3}\right)}}\\&={\sf{-\left(2^{3+3}\right)}}\\&={\sf{-\left(2^6\right)}}\\&={\sf{-64}}\end{aligned}

 

\begin{aligned}{\sf{c.}}~&{\sf{\frac{a^3b^{-2}c^2}{a^2b^{-1}c^4}}}\\&={\sf{\frac{5^3\cdot\left(-\frac{1}{4}\right)^{-2}\cdot2^2}{5^2 \cdot\left(-\frac{1}{4}\right)^{-1}\cdot2^4}}}\\&={\sf{\frac{5^3\cdot\left(\frac{1}{2^2}\right)^{-2}\cdot2^2}{5^2 \cdot\left(-\frac{1}{2^2}\right)^{-1}\cdot2^4}}}\\&={\sf{-\frac{5^3\cdot\left(2^2\right)^{-2}\cdot2^2}{5^2 \cdot\left(2^2\right)^{-1}\cdot2^4}}}\\&={\sf{-\frac{5^3\cdot2^{-4}\cdot2^2}{5^2 \cdot2^{-2}\cdot2^4}}}\\&={\sf{-\frac{5^3\cdot2^{-4+2}}{5^2 \cdot2^{-2+4}}}}\\&={\sf{-\frac{5^3\cdot2^{-2}}{5^2 \cdot2^{2}}}}\\&={\sf{-\left(5^{3-2}\cdot2^{-2-2}\right)}}\\&={\sf{-\left(5\cdot2^0\right)}}\\&={\sf{-\left(5\cdot1\right)}}\\&={\sf{-5}}\end{aligned}

 

\begin{aligned}{\sf{d.}}~&{\sf{\frac{ab^{-3}c^{-4}+a^2bc^3}{abc^2}}}\\&={\sf{\frac{5\cdot\left(-\frac{1}{4}\right)^{-3}\cdot2^{-4}+5^2\cdot\left(-\frac{1}{4}\right)\cdot2^3}{5\cdot\left(-\frac{1}{4}\right)\cdot2^2}}}\\&={\sf{\frac{5\cdot\left(- \left(\frac{1}{4}\right)^{-3}\right)\cdot2^{-4}+5^2\cdot\left(-\frac{1}{4}\right)\cdot2^3}{5\cdot\left(-\frac{1}{4}\right)\cdot2^2}}}\\&={\sf{\frac{-5\cdot\left(\frac{1}{2^2}\right)^{-3}\cdot2^{-4}-5^2\cdot\left(\frac{1}{2^2}\right)\cdot2^3}{5\cdot\left(-\frac{1}{\cancel{2^2}}\right)\cdot\cancel{2^2}}}}\\&={\sf{\frac{-5\cdot\left(2^{-2}\right)^{-3}\cdot2^{-4}-5^2\cdot\left(2^{-2}\right)\cdot2^3}{-5}}}\\&={\sf{\frac{-5\cdot2^6\cdot2^{-4}-5^2\cdot2^{-2}\cdot2^3}{-5}}}\\&={\sf{\frac{-5\cdot2^{6-4}-5^2\cdot2^{-2+3}}{-5}}}\\&={\sf{\frac{-5\cdot2^2-5^2\cdot2}{-5}}}\\&={\sf{\frac{-5\cdot4-25\cdot2}{5}}}\\&={\sf{\frac{-20-50}{-5}}}\\&={\sf{\frac{-70}{-5}}}\\&={\sf{14}}\end{aligned}

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Last Update: Tue, 15 Nov 22
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