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July 2022 2 11 Report

найдите сумму натуральных решений неравенства

[tex]\frac{x-2}{2}-\frac{3(2-x)}{10}+\frac{7x+1}{4}\leq\frac{x+11}{3}+\frac{13+16x}{20}[/tex]

irkarom

$\frac{x-2}{2}-\frac{3(2-x)}{10}+\frac{7x+1}{4}\leq\frac{x+11}{3}+\frac{13+16x}{20} \ |*60\\ 30(x-2)-6*3(2-x)+15(7x+1)\leq20(x+11)+3(13+16x)\\ 30x-60-36+18x+105x+15\leq20x+220+39+48x\\ 153x-81\leq68x+259\\ 153x-68x\leq259+81\\ 85x\leq340\\ x\leq4$

Ответ: (-∞; 4]

0 votes Thanks 1

Yena

$\frac{x-2}{2}-\frac{3(2-x)}{10}+\frac{7x+1}{4}\leq\frac{x+11}{3}+\frac{13+16x}{20} \\\ \frac{10(x-2)-6(2-x)+5(7x+1)}{20}\leq\frac{20(x+11)+3(13+16x)}{60} \\\ \frac{10x-20-12+6x+35x+5}{20}\leq\frac{20x+220+39+48x}{60} \\\ \frac{51x-27}{20}\leq\frac{68x+259}{60} \\\ \frac{3(51x-27)}{60}\leq\frac{68x+259}{60} \\\ 153x-81\leq68x+259 \\\153x-68x\leq259+81\\\ 85x\leq340\\ x\leq340:85\\\ x\leq4$

Натуральные решения этого неравенства будут 1,2,3,4

найдем их сумму 1+2+3+4=10

1 votes Thanks 1

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