\(-9\cdot4^{\frac{1}{x}}-5\cdot6^{\frac{1}{x}}+4\cdot9^{\frac{1}{x}}=0\)
LH
Những câu hỏi liên quan
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
Đúng 0
Bình luận (0)
\(\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-\frac{1}{7\cdot6}-\frac{1}{6\cdot5}-\frac{1}{5\cdot4}-\frac{1}{4\cdot3}-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
Đúng 0
Bình luận (0)
\(41\sqrt[9^1]{8\sqrt[2]{\frac{12}{2.85\frac{1\cdot2+3\cdot4+5\cdot6+7\cdot8+9\sqrt[4]{16}}{2\cdot\frac{12}{2}\sqrt{4^2}-7^2}}}4\cdot5\cdot6\cdot7\cdot8\cdot9}\)
Ô phép tính khủng. Cái này do bạn chế ra à !
Đúng 0
Bình luận (0)
\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
Rút gọn:
a,\(A=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
b,\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2014\cdot2016}\right)\)
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot18}=...\)
Nhập kết quả dưới dạng phân số tối giản
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)\(\frac{1}{18}\)
= \(\frac{1}{3}-\frac{1}{18}\)
= \(\frac{5}{18}\)
Đúng 0
Bình luận (0)
[{1/4-1/4}+ { 1/6-1/6}+{1/9-1/9}+{1/13-1/13}]*11/4=0
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH :
\(\frac{2}{5}< A< \frac{8}{9}\)
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)
Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)
Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)
Bài làm :
Ta có :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}\)
\(A>\frac{2}{5}\left(1\right)\)
Ta cũng có :
\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}\)
\(A< \frac{8}{9}\left(2\right)\)
\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(S=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
Bài làm:
Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)
Cái còn lại tự CM
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\frac{1}{8\cdot8}+\frac{1}{9\cdot9}\)
HÃY CHỨNG MINH \(\frac{2}{5}< S< \frac{7}{8}\)
A= 1/2.2 + 1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9
Vì 1/2.2 > 1/2.3; 1/3.3 > 1/3.4 ; 1/5.5 > 1/5.6;...... nên
1/2.2 +1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9 > 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
Ta có: 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
= 1/2-1/3 + 1/3 -1/4 + 1/4-1/5+...+1/9-1/10
= 1/2- 1/10
= 2/5
Vì A < 2/5 mà 2/5 <7/8 nên 2/5 < A < 7/8
Vậy....