x uchun yechish
x\leq 19
Grafik
Viktorina
Algebra
5xshash muammolar:
5 ( \frac { x } { 5 } + \frac { 10 } { 2 } ) \geq 2 x + \frac { 30 } { 5 }
Baham ko'rish
Klipbordga nusxa olish
50\left(\frac{x}{5}+\frac{10}{2}\right)\geq 20x+2\times 30
Tenglamaning ikkala tarafini 10 ga, 5,2 ning eng kichik karralisiga ko‘paytiring. 10 >0 bo‘lganda, tengsizlikning yo‘nalishi o‘zgarmasdan qoladi.
50\left(\frac{x}{5}+5\right)\geq 20x+2\times 30
5 ni olish uchun 10 ni 2 ga bo‘ling.
50\times \frac{x}{5}+250\geq 20x+2\times 30
50 ga \frac{x}{5}+5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.
10x+250\geq 20x+2\times 30
50 va 5 ichida eng katta umumiy 5 faktorini bekor qiling.
10x+250\geq 20x+60
60 hosil qilish uchun 2 va 30 ni ko'paytirish.
10x+250-20x\geq 60
Ikkala tarafdan 20x ni ayirish.
-10x+250\geq 60
-10x ni olish uchun 10x va -20x ni birlashtirish.
-10x\geq 60-250
Ikkala tarafdan 250 ni ayirish.
-10x\geq -190
-190 olish uchun 60 dan 250 ni ayirish.
x\leq \frac{-190}{-10}
Ikki tarafini -10 ga bo‘ling. -10 <0 bo‘lganda, tengsizlikning yo‘nalishi o‘zgargan bo‘ladi.
x\leq 19
19 ni olish uchun -190 ni -10 ga bo‘ling.
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