Trang chủ » Hỏi đáp » Môn Toán Cho B= 3^3/10.13+ 3^3/13.16+ 3^3/16.19+…+ 3^3/67.70 CMR B<1 29/08/2023 Cho B= 3^3/10.13+ 3^3/13.16+ 3^3/16.19+…+ 3^3/67.70 CMR B<1
B=3^3/10.13 +3^3/13.16 +3^3/16.19+ ….+ 3^3/67.70 suy ra B= 3^3/3 . ( 3/10.13+3/13.16 + ….+ 3/67.70 ) suy ra B=9.(1/10-1/13+1/13-1/16+…+1/67-1/70) suy ra B= 9.( 1/10-1/70) suy ra B=9. 3/25 suy ra B=27/25 <1 ( điều cần chứng minh ) Trả lời
B = 3^3/(10.13) + 3^3/(13.16) + … + 3^3/(67.70)=>B = 3^3/3 (3/(10.13) + 3/(13.16) + … + 3/(67.70))=>B = 9(1/10 – 1/13 + 1/13 – 1/16 + … + 1/67 – 1/70)=>B = 9(1/10 – 1/70)=>B = 9 . 3/35 = 27/35 < 1 (đpcm) Trả lời
=>B = 3^3/3 (3/(10.13) + 3/(13.16) + … + 3/(67.70))
=>B = 9(1/10 – 1/13 + 1/13 – 1/16 + … + 1/67 – 1/70)
=>B = 9(1/10 – 1/70)
=>B = 9 . 3/35 = 27/35 < 1 (đpcm)