(12,2-7 6/11): 3ц 7/33 (5/9+7ц 7/8*2ц 1/7)*4ц 3/3 - вопрос №50035241227611373359778217433 от slyusarevqaowko3q 04.02.2023 08:48

Question

Математика: (12,2-7 6/11): 3ц 7/33 (5/9+7ц 7/8*2ц 1/7)*4ц 3/3 решите

slyusarevqaowko3q · Answer

(12, 2 - 7ц 6/11) / (3ц 7/33)= (12ц 22/110 - 7ц 60/110) /(3ц 7/33) = 4ц 72/110 /(3ц 7/33) = 4ц 36/55 /(3ц 7/33) =(256/55) / (106/33)= (256/55) * (33/106) = (128*3) / (5*53) = 384/265 = 1ц 119/265

(5/9 + 7ц7/8 *2ц1/7) * 4ц3/3 = (5/9 + 63/8*15/7) *5 = (5/9 + (9*15)/(8*1)) *5 =(5/9 + 135/8) *5 = (40/72 + 1215/72) *5 = 1255/72 *5 = 1255*5/72= 6275 /72= 87ц 11/75