1.

Prove that : (i)(√3×5−3÷3√3−1√5)×6√3×56=35(ii)932−3×50−(181)−12=15(iii)(14)−2−3×823×40+(916)−12=163(iv)212×313×41410−15×535÷343×5−754−35×6=10(v)√14+(0.01)−12−(27)23=32(vi)2n+2n−12n+1−2n=32(vii)(64125)−23+1(256625)14+(√253√64)0=6116(viii)3−3×62×√9852×3√125×(15)−43×313=28√2(ix)(0.6)0−(0.1)−1(38)−1(32)3+(13)−1=−32

Answer»

Prove that :

(i)(3×53÷3315)×63×56=35(ii)9323×50(181)12=15(iii)(14)23×823×40+(916)12=163(iv)212×313×4141015×535÷343×575435×6=10(v)14+(0.01)12(27)23=32(vi)2n+2n12n+12n=32(vii)(64125)23+1(256625)14+(25364)0=6116(viii)33×62×9852×3125×(15)43×313=282(ix)(0.6)0(0.1)1(38)1(32)3+(13)1=32



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