Question

Решить уравнения 45 1) 3,4*(5х-0,16)=0,4762)128: (2,4х+19,04)=6,43)4,4: х: 0,3=0,264)11,2: y+0,32=0,65)(x+9,14): 7,2=5

Answer 1

$\displaystyle\tt 1)\\ 3,4(5x-0,16)=0,476\\5x-0,16=0,476:3,4\\5x-0,16=0,14\\5x=0,14+0,16\\5x=0,3\\x=0,3:5\\x=0,06\\\\ 2)\\ 128:(2,4x+19,04) = 6,4\\2,4x + 19,04 = 128 : 6,4\\2,4x + 19,04 = 20\\2,4x = 20 - 19,04\\ 2,4x = 0,96\\ x = 0,96 : 2,4\\x = 0,4\\\\$

$\displaystyle\tt3)\\4,4:x:0,3=0,26\\\\ \frac{44}{10}\cdot\frac{1}{x}\cdot\frac{10}{3}=\frac{26}{100}\\\\ \frac{44}{3x}=\frac{13}{50}; \ \ \ x\neq0\\\\ 3x\cdot13=44\cdot50\\\\ 39x=2200\\\\x=\frac{2200}{39}=56\frac{16}{39}$

$\displaystyle\tt 4)\\ 11,2:y+0,32=0,6\\ 11,2:y=0,6-0,32\\11,2:y=0,28\\ y=11,2:0,28\\y=40\\\\ 5)\\ (x+9,14):7,2=5\\x+9,14=5\cdot7,2\\x+9,14=36\\x=36-9,14\\x=26,86$