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*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
Bài 1:
b) Ta có: \(D=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)
\(=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot0\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)
=0
Bài làm
\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Rightarrow\frac{22}{30}-\frac{27}{30}< x< \frac{11}{15}.\frac{10}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{11}{3}.\frac{2}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{22}{27}\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
~ Chắc z ~
# Chúc bạn học tốt #
Ta có:\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Leftrightarrow\frac{110-135}{30}< x< \frac{11.10}{15.9}\)
\(\Leftrightarrow\frac{-15}{30}< x< \frac{22}{27}\)
(Vì x c Z)\(\Leftrightarrow-1< x< 1\Rightarrow x\in\left\{0\right\}\)
(x - 1)x + 2 = (x - 1)x + 6
(x - 1)x + 2 - (x - 1)x + 6 = 0
(x - 1)x + 2.[1 - (x - 1)x + 4] = 0
\(\Rightarrow\) (x - 1)x + 2 = 0 hoặc 1 - (x - 1)x + 4 = 0
\(\Rightarrow\) x - 1 = 0 hoặc (x - 1)x + 4 = 1
\(\Rightarrow\) x = 1 hoặc x - 1 = 1 hoặc x - 1 = - 1
\(\Rightarrow\) x = 1 hoặc x = 2 hoặc x = 0
Vậy \(x\in\left\{0;1;2\right\}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có
\(\frac{x_1-1}{10}=.....=\frac{x_{10}-10}{1}=\frac{\left(x_1+x_2+....+x_{10}\right)-\left(1+2+3+...+10\right)}{1+2+3+...+10}\)
\(=\frac{45}{55}=\frac{9}{11}\)
Giải ra ta được
\(x_1=\frac{101}{11}\)
\(x_2=\frac{103}{11}\)
........
\(x_{10}=\frac{119}{11}\)
1/4×2/6×3/8×4/10×...×14/30×15/32=1/2^x
<=>1/(2×2)×2/(2×3)×...×14/(2×15)×15/2^5=1/2^x
<=>1/2×1/2×...×1/2×1/(2^5)=1/2^x
<=>1/2^19=1/2^x=>x=19
Đề mình không ghi lại nhé.
\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{4\times6\times10\times...\times30\times32}=\frac{1}{2^x}\)\(\frac{1}{2^x}\)
\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{2\times4\times6\times8\times10\times...\times30\times32}\)\(=\frac{1}{2^{x+1}}\)
\(\Rightarrow\frac{1}{2^{15}\times32}=\)\(\frac{1}{2^{x+1}}\)
\(\Rightarrow2^{15}\times2^5=2^{x+1}\)
\(\Rightarrow2^{20}=2^{x+1}\)
\(\Rightarrow x+1=20\Rightarrow x=19\)
Vậy \(x=1\)
Học tốt nhaaa!