Question

Номер 1034 Решите неравенство:

Номер 1039 Вычислите:

Answer 1

$\frac{\left(2\frac34+\frac58\right)\cdot\frac49}{\left(8\frac16-3\frac12\right):1\frac16}+\frac{\left(4,8-1\frac25\right):0,68}{29:\left(2\frac12+1\frac18\right)}\\ \left(2\frac34+\frac58\right)\cdot\frac49=\left(\frac{11}4+\frac58\right)\cdot\frac49=\frac{27}8\cdot\frac49=\frac32\\ \left(8\frac16-3\frac12\right):1\frac16=\left(\frac{49}6-\frac72\right):\frac76=\frac{28}6\cdot\frac67=4$

$\frac{\left(2\frac34+\frac58\right)\cdot\frac49}{\left(8\frac16-3\frac12\right):1\frac16}=\frac32:4=\frac32\cdot\frac14=\frac3{12}$

$\left(4,8-1\frac25\right):0,68=\left(\frac{48}{10}-\frac75\right):\frac{68}{100}=\frac{34}{10}\cdot\frac{100}{68}=\frac{10}2=5\\ 29:\left(2\frac12+1\frac18\right)=29:\left(\frac52+\frac98\right)=29:\frac{29}8=29\cdot\frac8{29}=8\\ \frac{\left(4,8-1\frac25\right):0,68}{29:\left(2\frac12+1\frac18\right)}=\frac58$

$\frac{\left(2\frac34+\frac58\right)\cdot\frac49}{\left(8\frac16-3\frac12\right):1\frac16}+\frac{\left(4,8-1\frac25\right):0,68}{29:\left(2\frac12+1\frac18\right)}=\frac3{12}+\frac58=\frac{12+30}{48}=\frac{42}{48}=\frac78$

$1033.1\quad9-\frac{5x+1}2>x+5\quad\times2\\ 18-5x-1>2x+10\\ -5x-2x>10-17\\ -7x>-7\\ x<1\\ 2.\quad1,75+\frac{2x}3\frac14\\ 3.\quad\frac{4+y}2-\frac{y+2}7-2\\ 4.\quad4+\frac{7y-3}{5}>\frac{3y+5}4-\frac{3y}2\quad\times20\\ 80+28y-12>15y+25-30y\\ 28y+15y>25-68\\ 43y>-43\\ y>-1$

$1034.1\quad\frac{7x+2}6-x\leq\frac{5x+4}3-4x\quad\times6\\ 7x+2-6x\leq10x+8-24x\\ x+14x\leq8-2\\ 15x\leq6\\ x\leq\frac35\\ 2.\quad\frac{4x+1}3-x>\frac{x+1}2-\frac{x-3}4\quad\times12\\ 16x+4-12x>6x+6-3x+9\\ 4x-3x>15-4\\ x>11\\ 3.\quad\frac{9-5x}2-\frac{4x}31\\ 4.\quad\frac{5x-3}4-\frac{11x}6\geq\frac{2(1-x)}3+\frac{3x}2\quad\times12\\ 15x-9-22x\geq8-8x+18x\\ -7x-10x\geq8+9\\ -17x\geq17\\ x\leq-1$