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Bilangan bulat terkecil yang memenuhi pertidaksamaan 4 3 x − 4 ​ < 2 4 ( x − 1 ) ​ adalah ...

Bilangan bulat terkecil yang memenuhi pertidaksamaan  adalah ... 

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D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

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bilangan bulat terkecil x yang memenuhi adalah 1 .

bilangan bulat terkecil   yang memenuhi adalah 

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Pembahasan

Perhatikan perhitungan berikut! &\frac{-4}{{\color[rgb]{0.0, 0.0, 1.0}-}{\color[rgb]{0.0, 0.0, 1.0}5}}\\x&>&\frac45\end{array}" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mn»3«/mn»«mi»x«/mi»«mo»-«/mo»«mn»4«/mn»«/mrow»«mn»4«/mn»«/mfrac»«/mtd»«mtd»«mo»§#60;«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»4«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/mrow»«mn»2«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfenced»«mfrac»«mrow»«mn»3«/mn»«mi»x«/mi»«mo»-«/mo»«mn»4«/mn»«/mrow»«mn»4«/mn»«/mfrac»«/mfenced»«mo»§#183;«/mo»«mn mathcolor=¨#0000FF¨»4«/mn»«/mtd»«mtd»«mo»§#60;«/mo»«/mtd»«mtd»«mfenced open=¨[¨ close=¨]¨»«mfrac»«mrow»«mn»4«/mn»«mfenced»«mrow»«mi»x«/mi»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mfenced»«/mrow»«mn»2«/mn»«/mfrac»«/mfenced»«mo»§#183;«/mo»«mn 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mathcolor=¨#0000FF¨»x«/mi»«/mtd»«/mtr»«mtr»«mtd»«mo»-«/mo»«mn»5«/mn»«mi»x«/mi»«/mtd»«mtd»«mo»§#60;«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»4«/mn»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mo»-«/mo»«mn»5«/mn»«mi»x«/mi»«/mrow»«mrow»«mo mathcolor=¨#0000FF¨»-«/mo»«mn mathcolor=¨#0000FF¨»5«/mn»«/mrow»«/mfrac»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mfrac»«mrow»«mo»-«/mo»«mn»4«/mn»«/mrow»«mrow»«mo mathcolor=¨#0000FF¨»-«/mo»«mn mathcolor=¨#0000FF¨»5«/mn»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mfrac»«mn»4«/mn»«mn»5«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»" role="math" 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style="max-width: none;"> Dengan demikian, bilangan bulat terkecil x yang memenuhi adalah 1 .

Perhatikan perhitungan berikut! 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 3 x minus 4 over denominator 4 end fraction end cell less than cell fraction numerator 4 open parentheses x minus 1 close parentheses over denominator 2 end fraction end cell row cell open parentheses fraction numerator 3 x minus 4 over denominator 4 end fraction close parentheses times 4 end cell less than cell open square brackets fraction numerator 4 open parentheses x minus 1 close parentheses over denominator 2 end fraction close square brackets times 4 end cell row cell 3 x minus 4 end cell less than cell 8 open parentheses x minus 1 close parentheses end cell row cell 3 x minus 4 end cell less than cell 8 x minus 8 end cell row cell 3 x minus 4 plus 4 end cell less than cell 8 x minus 8 plus 4 end cell row cell 3 x end cell less than cell 8 x minus 4 end cell row cell 3 x minus 8 x end cell less than cell 8 x minus 4 minus 8 x end cell row cell negative 5 x end cell less than cell negative 4 end cell row cell fraction numerator negative 5 x over denominator negative 5 end fraction end cell greater than cell fraction numerator negative 4 over denominator negative 5 end fraction end cell row x greater than cell 4 over 5 end cell end table  

Dengan demikian, bilangan bulat terkecil   yang memenuhi adalah 

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Penyelesaian dari pertidaksamaan 3 2 x − 3 ​ − 2 3 x − 4 ​ ≤ 5 adalah ...

95

4.8

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