Let me cook From the problem we can multiply the dominator with (√40+√35) as we know (a^2-b^2)=(a-b)(a+b) ----> 1/(40-35)×(√40+√35)^2 ----> 1/5×(√40+√35)^2 ----> 1/5×(√40^2+√35^2+2√(40×35) ----> 1/5)×40+35+2×√1400 ----> 1/5×(75+2(√100×√14) ----> 1/5((75+2(10√14) ----> 1/5(75+20√14) ----> 15+4√14
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