Tính nhanh : \(2015^{\left(196-1^2\right).\left(196-2^2\right)....\left(196-14^2\right)}\)
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12 tháng 5 2021
mình trả lời rồi đó :
,hzgb3fewniurse8w23uhxu9ew8ahbg18yhr7tgfse7w6hb ch
TG
21 tháng 4 2019
Bài 1:
Vì \(\frac{196}{197+198}< \frac{196}{197};\frac{197}{197+198}< \frac{197}{198}\)
Nên A = \(\frac{196}{197}+\frac{197}{198}>\frac{196}{197+198}+\frac{197}{197+198}=\frac{196+197}{197+198}=B\)
Vậy A > B
TA
1
1 tháng 8 2019
nhân 2 vế cho 12 ta được
(6x+5)2(3x+2)(12x+12)=2352
bạn nhóm cho có phần chung r làm như mấy bài trước nha :))
2 tháng 2 2022
\(M=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=\dfrac{84-30-5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=\dfrac{49}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=\dfrac{7}{12}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=\dfrac{7\cdot17-5}{204}-\dfrac{5}{374}=\dfrac{19}{34}-\dfrac{5}{374}=\dfrac{204}{374}=\dfrac{6}{11}\)
31 tháng 12 2023
a: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{1}{2}\left(2+3+4+...+17\right)\)
\(=\dfrac{1}{2}\cdot152=76\)
b: Sửa đề: \(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right)\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{193}\cdot\dfrac{193}{17}-\dfrac{3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left[\dfrac{7}{1931}\cdot\dfrac{1931}{25}+\dfrac{11}{3862}\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)
\(=\left(\dfrac{2}{17}+\dfrac{30}{34}\right):\dfrac{14+11+225}{50}\)
\(=1\cdot\dfrac{50}{250}=1\cdot\dfrac{1}{5}=\dfrac{1}{5}\)
c: Sửa đề: \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
d: \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)
\(=\dfrac{1}{3}+1:\dfrac{3}{2}=1\)
24 tháng 12 2023
a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)
\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)
\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)
\(=2015^{\left(196-1^2\right).\left(196-2^2\right)....\left(196-196\right)}\)
\(=2015^{\left(196-1^2\right).\left(196-2^2\right)....0}\)
\(=2015^0=1\)