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

Tolong kaaaa dua soal ini

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\mathbb \color{aqua} \underbrace{JAWABAN}$

No. 19 :

$\small \boxed{ \bf {f}^{ - 1} (5) = - 5 \frac{6}{7} } \\$

No. 20 :

$\small \boxed{ \bf {g}^{ - 1} (3) = 2 \frac{1}{3} }$

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No. 19 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

$\tt f(x) = \frac{3x - 6}{2x + 7}$

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\tt {f}^{ - 1} (5)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

⟹ menentukan f-¹(x) :

$\begin{aligned} \tt y &= \tt \frac{3x - 6}{2x + 7} \\ \tt y(2x + 7) &= \tt 3x - 6 \\ \tt 2xy + 7y &= \tt 3x - 6 \\ \tt 2xy - 3x &= \tt - 7y - 6 \\ \tt x(2y - 3) &= \tt - 7y - 6 \\ \tt x &= \tt \frac{ - 7y - 6}{2y - 3} \end{aligned}$

$\tt {f}^{ - 1} (x) = \dfrac{ - 7x - 6}{2x - 3} \\$

⟹ menentukan f-¹(5) :

$\begin{aligned} \tt {f}^{ - 1} (5) &= \tt \frac{ - 7(5) - 6}{2(5) - 3} \\ \tt &= \tt \frac{ - 35 - 6}{10 - 3} \\ &= \tt \frac{ - 41}{7} \\ &= \bf - 5 \frac{6}{7} \end{aligned} \\$

No. 20 :

$\underline{ \overline{ \boxed{ \bold{diketahui}}}}$

$\tt g(x) = \frac{3x}{4x - 7}$

$\\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}$

$\tt {g}^{ - 1} (3)$

$\\ \underline{ \overline{ \boxed{ \bold{jawab}}}}$

⟹ menentukan g-¹(x) :

$\begin{aligned} \tt y &= \tt \frac{3x}{4x - 7} \\ \tt y(4x - 7) &= \tt 3x \\ \tt 4xy - 7y &= \tt 3x \\ \tt 4xy - 3x &= \tt 7y \\ \tt x(4y - 3) &= \tt 7y \\ \tt x &= \tt \frac{7y}{4y - 3} \end{aligned} \\$

$\tt {g}^{ - 1} (x) = \frac{7x}{4x - 3} \\$

⟹ menentukan g-¹(3) :

$\begin{aligned} \tt {g}^{ - 1} (3) &= \tt \frac{7(3)}{4(3) - 3} \\ &= \tt \frac{21}{12 - 3} \\ &= \tt \frac{21}{9} \\ &= \tt 2 \frac{3}{9} \\ &= \bf 2 \frac{1}{3} \end{aligned}$

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$\mathbb \color{red} \underbrace{KESIMPULAN}$

$\text{[math]}$ Jadi, nilai $\bf {f}^{ - 1} (5)$ adalah $\bf - 5 \frac{6}{7}$ dan nilai $\bf {g}^{ - 1} (3)$ adalah $\bf 2 \frac{1}{3}$

$\colorbox{ff0000}{} \colorbox{ff4000}{}\colorbox{ff8000}{}\colorbox{ffc000}{}\colorbox{ffff00}{}\colorbox{c0ff00}{}\colorbox{80ff00}{}\colorbox{40ff00}{}\colorbox{00ff00}{}\colorbox{00ff40}{}\colorbox{00ff80}{}\colorbox{00ffc0}{}\colorbox{00ffff}{}\colorbox{00c0ff}{}\colorbox{0080ff}{}\colorbox{0040ff}{}\colorbox{0000ff}{}\colorbox{4000ff}{}\colorbox{8000ff}{}\colorbox{c000ff}{}\colorbox{ff00ff}{}\colorbox{ff00c0}{}\colorbox{ff00a0}{}\colorbox{ff0080}{}\colorbox{ff0040}{}$

$[tex] \mathbb \color{aqua} \underbrace{JAWABAN}[/tex]No. 19 :[tex] \small \boxed{ \bf {f}^{ - 1} (5) = - 5 \frac{6}{7} } \\ [/tex]No. 20 : [tex] \small \boxed{ \bf {g}^{ - 1} (3) = 2 \frac{1}{3} } [/tex]------------------No. 19 :[tex] \underline{ \overline{ \boxed{ \bold{diketahui}}}}[/tex][tex] \tt f(x) = \frac{3x - 6}{2x + 7} [/tex][tex] \\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}[/tex][tex] \tt {f}^{ - 1} (5)[/tex][tex] \\ \underline{ \overline{ \boxed{ \bold{jawab}}}}[/tex]⟹ menentukan f-¹(x) : [tex] \begin{aligned} \tt y &= \tt \frac{3x - 6}{2x + 7} \\ \tt y(2x + 7) &= \tt 3x - 6 \\ \tt 2xy + 7y &= \tt 3x - 6 \\ \tt 2xy - 3x &= \tt - 7y - 6 \\ \tt x(2y - 3) &= \tt - 7y - 6 \\ \tt x &= \tt \frac{ - 7y - 6}{2y - 3} \end{aligned}[/tex][tex] \tt {f}^{ - 1} (x) = \dfrac{ - 7x - 6}{2x - 3} \\ [/tex]⟹ menentukan f-¹(5) : [tex] \begin{aligned} \tt {f}^{ - 1} (5) &= \tt \frac{ - 7(5) - 6}{2(5) - 3} \\ \tt &= \tt \frac{ - 35 - 6}{10 - 3} \\ &= \tt \frac{ - 41}{7} \\ &= \bf - 5 \frac{6}{7} \end{aligned} \\ [/tex]No. 20 :[tex] \underline{ \overline{ \boxed{ \bold{diketahui}}}}[/tex][tex] \tt g(x) = \frac{3x}{4x - 7} [/tex][tex] \\ \underline{ \overline{ \boxed{ \bold{ditanya}}}}[/tex][tex] \tt {g}^{ - 1} (3)[/tex][tex] \\ \underline{ \overline{ \boxed{ \bold{jawab}}}}[/tex]⟹ menentukan g-¹(x) : [tex] \begin{aligned} \tt y &= \tt \frac{3x}{4x - 7} \\ \tt y(4x - 7) &= \tt 3x \\ \tt 4xy - 7y &= \tt 3x \\ \tt 4xy - 3x &= \tt 7y \\ \tt x(4y - 3) &= \tt 7y \\ \tt x &= \tt \frac{7y}{4y - 3} \end{aligned} \\ [/tex][tex] \tt {g}^{ - 1} (x) = \frac{7x}{4x - 3} \\ [/tex]⟹ menentukan g-¹(3) : [tex] \begin{aligned} \tt {g}^{ - 1} (3) &= \tt \frac{7(3)}{4(3) - 3} \\ &= \tt \frac{21}{12 - 3} \\ &= \tt \frac{21}{9} \\ &= \tt 2 \frac{3}{9} \\ &= \bf 2 \frac{1}{3} \end{aligned}[/tex]------------------[tex] \mathbb \color{red} \underbrace{KESIMPULAN}[/tex][tex] [/tex]Jadi, nilai [tex] \bf {f}^{ - 1} (5) [/tex] adalah [tex] \bf - 5 \frac{6}{7} [/tex] dan nilai [tex] \bf {g}^{ - 1} (3) [/tex] adalah [tex] \bf 2 \frac{1}{3} [/tex][tex] \colorbox{ff0000}{} \colorbox{ff4000}{}\colorbox{ff8000}{}\colorbox{ffc000}{}\colorbox{ffff00}{}\colorbox{c0ff00}{}\colorbox{80ff00}{}\colorbox{40ff00}{}\colorbox{00ff00}{}\colorbox{00ff40}{}\colorbox{00ff80}{}\colorbox{00ffc0}{}\colorbox{00ffff}{}\colorbox{00c0ff}{}\colorbox{0080ff}{}\colorbox{0040ff}{}\colorbox{0000ff}{}\colorbox{4000ff}{}\colorbox{8000ff}{}\colorbox{c000ff}{}\colorbox{ff00ff}{}\colorbox{ff00c0}{}\colorbox{ff00a0}{}\colorbox{ff0080}{}\colorbox{ff0040}{} [/tex]$