Предмет: Математика, автор: kandybko93

Номер 977
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kandybko93: Это 6 класс, авторы Т.А. Алдамуратова; К.С. Байшоланова...

Ответы

Автор ответа: ksyuwi
1

( 1 )

\left \{ {{\frac{x}{6}+\frac{x}{3} <2 } \atop {2-\frac{1}{3}x>0 }} \right. \\\\\left \{ {{\frac{1}{6}x +\frac{2}{6}x<2 } \atop {-\frac{1}{3}x>-2 }} \right. \\\\\left \{ {{\frac{3}{6}x<2 } \atop {x<-2:(-\frac{1}{3}) }} \right. \\\\\left \{ {{x<2:\frac{3}{6} } \atop {x<-2*(-\frac{3}{1}) }} \right. \\\\\left \{ {{x<2*\frac{6}{3} } \atop {x<\frac{6}{1}  }} \right.\\\\\left \{ {{x<\frac{12}{3} } \atop {x<6 }} \right.\\\\\left \{ {{x<4 } \atop {x<6  }} \right

x ∈ ( -∞ ; 4 )

( 2 )

\left \{ {{x-\frac{x+3}{2}\geq 1 } \atop {-\frac{x}{2}\leq 2-\frac{x}{3}  }} \right. \\\\\left\{{{\frac{2x}{2}-\frac{x}{2}+\frac{3}{2}\geq1}\atop{-\frac{x}{2}+\frac{x}{3}\leq2}} \right. \\\\\left \{ {{\frac{x}{2}\geq 1-\frac{3}{2}} \atop {-\frac{x}{2}+\frac{x}{3} \leq 2  }} \right. \\\\\left \{ {{\frac{x}{2}\geq \frac{2}{2} -\frac{3}{2}} \atop {-\frac{3x}{6}+\frac{2x}{6} \leq 2  }} \right.\\\\\left \{ {{\frac{x}{2}\geq-\frac{1}{2}} \atop {-\frac{x}{6} \leq 2  }} \right.\\

\left \{ {{x\geq -\frac{1}{2}:\frac{1}{2} } \atop {x\geq 2:(-\frac{1}{6})   }} \right.\\\\\left \{ {{x\geq -\frac{1}{2}*\frac{2}{1} } \atop {x\geq 2*(-\frac{6}{1})   }} \right.\\\\\left \{ {{x\geq -\frac{2}{2}} \atop {x\geq-\frac{2}{6}  }} \right.\\\\\left \{ {{x\geq -1} \atop {x\geq -\frac{1}{3} }} \right.

x ∈ ( -\frac{1}{3} ; + ∞ )

( 3 )

\left \{ {{\frac{x}{3} -\frac{x}{4} <\frac{x}{6} -1} \atop {6-\frac{x}{2} >\frac{x}{4}+3 }} \right. \\\\\left \{ {{\frac{1}{3}x -\frac{1}{4}x -\frac{1}{6}x < -1} \atop {-\frac{1}{2}x-\frac{1}{4}x>+3-6 }} \right.\\\\\left \{ {{\frac{4}{12}x -\frac{3}{12}x -\frac{2}{12}x < -1} \atop {-\frac{4}{8}x-\frac{2}{8}x>3-6 }} \right.\\\\\left \{ {{-\frac{1}{12}x < -1} \atop {-\frac{6}{8}x>-3}} \right.\\\\\left \{ {{x>-1:(-\frac{1}{12} )} \atop {x<-3:(-\frac{6}{8} )}} \right. \\

\left \{ {{x>-1*(-\frac{12}{1} )} \atop {x<-3*(-\frac{8}{6} )}} \right. \\\\\left \{ {{x>\frac{12}{1} } \atop {x<\frac{24}{6} }} \right. \\\\\left \{ {{x>12} \atop {x<4}} \right.

x ∈ ( -∞ ; 4 ) ∪ ( 12 ; +∞ )

( 4 )

\left \{ {{\frac{x}{5} -\frac{2}{3} <\frac{2}{5} -\frac{x}{3} } \atop {\frac{2}{7} +\frac{x}{3} >\frac{x}{7} -\frac{2}{3} }} \right. \\\\\left \{ {{\frac{x}{5} +\frac{x}{3} <\frac{2}{5} +\frac{2}{3} } \atop {\frac{x}{3} -\frac{x}{7} >-\frac{2}{3} -\frac{2}{7} }} \right.\\\\\left \{ {{\frac{3x}{15} +\frac{5x}{15} <\frac{6}{15} +\frac{10}{15} } \atop {\frac{7x}{21} -\frac{3x}{21} >-\frac{14}{21} -\frac{6}{21} }} \right.\\

\left \{ {{\frac{8x}{15}<\frac{16}{15}} \atop {\frac{4x}{21}>-\frac{20}{21}}} \right.\\\\\left \{ {{x<\frac{16}{15}:\frac{8}{15} } \atop {x>-\frac{20}{21}:\frac{4}{21} }} \right.\\\\\left \{ {{x<\frac{16}{15}*\frac{15}{8} } \atop {x>-\frac{20}{21}*\frac{21}{4} }} \right.\\\\\left \{ {{x<\frac{2}{1}*\frac{1}{1} } \atop {x>-\frac{5}{1}*\frac{1}{1} }} \right.\\\\\left \{ {{x<\frac{2}{1} } \atop {x>-\frac{5}{1} }} \right. \\\\\left \{ {{x<2} \atop {x>-5}} \right.

x ∈ ( -5 ; 2 )

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