[tex]\displaystyle\bf\\1)\\\\\frac{2}{9} \Big(1,8m-5,4\Big)-\frac{3}{7} \Big(2,1m-4,2\Big)=\frac{2}{9}\cdot 1,8m-\frac{2}{9} \cdot 5,4-\frac{3}{7} \cdot 2,1m+\frac{3}{7} \cdot 4,2=\\\\\\=2\cdot 0,2m-2\cdot0,6-3\cdot 0,3m+3\cdot 0,6=0,4m-1,2-0,9m+1,8=\\\\\\=0,6-0,5m\\\\\\2)\\\\\frac{1}{3} \Big(0,3y-0,6\Big)-\frac{1}{4} \Big(0,4y-0,8\Big)=\frac{1}{3}\cdot 0,3y-\frac{1}{3} \cdot 0,6-\frac{1}{4} \cdot 0,4y+\frac{1}{4} \cdot 0,8=\\\\\\=0,1y-0,2-0,1y+0,2=0[/tex]
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[tex]\displaystyle\bf\\1)\\\\\frac{2}{9} \Big(1,8m-5,4\Big)-\frac{3}{7} \Big(2,1m-4,2\Big)=\frac{2}{9}\cdot 1,8m-\frac{2}{9} \cdot 5,4-\frac{3}{7} \cdot 2,1m+\frac{3}{7} \cdot 4,2=\\\\\\=2\cdot 0,2m-2\cdot0,6-3\cdot 0,3m+3\cdot 0,6=0,4m-1,2-0,9m+1,8=\\\\\\=0,6-0,5m\\\\\\2)\\\\\frac{1}{3} \Big(0,3y-0,6\Big)-\frac{1}{4} \Big(0,4y-0,8\Big)=\frac{1}{3}\cdot 0,3y-\frac{1}{3} \cdot 0,6-\frac{1}{4} \cdot 0,4y+\frac{1}{4} \cdot 0,8=\\\\\\=0,1y-0,2-0,1y+0,2=0[/tex]