24⋮x,15⋮(x+1)
K
Khách
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2 tháng 3 2022
\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)
N
0
26 tháng 3 2017
=1/(1x3)x1/(2x4)x...x1/(9x11)
=1/(1x3x2x4x...x9x11)
=1/(1x2x3x3x4x4x5x5x...x9x10x11)
R
8 tháng 7 2019
a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
TH
0
PT
1
CM
7 tháng 5 2017
a) 47 – x + 15 = 21 ⇔ 47 – x − 15 = 21 ⇔ 47 − x = 21 + 15 ⇔ 7 − x = 36 ⇔ − x = 47 − 36 ⇔ − x = 11 ⇔ x = − 11.
b) – 5 – 24 – x = – 11 ⇔ − 5 − 24 + x = − 11 ⇔ − 29 + x = − 11 ⇔ x = − 11 + 29 ⇔ x = 18.
c) 11 + 15 – x = 1 ⇔ 15 − x = 1 − 11 ⇔ 15 − x = − 10 ⇔ 15 + 10 = x ⇔ x = 25.
ND
1
21 tháng 10 2023
\(K=\left(1-\dfrac{3}{2\cdot4}\right)\left(1-\dfrac{3}{3\cdot5}\right)\cdot...\cdot\left(1-\dfrac{3}{19\cdot21}\right)\)
\(=\dfrac{3^2-1-3}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{4^2-1-3}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{20^2-4}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{\left(3-2\right)\left(3+2\right)}{\left(3-1\right)\left(3+1\right)}\cdot\dfrac{\left(4-2\right)\left(4+2\right)}{\left(4-1\right)\left(4+1\right)}\cdot...\cdot\dfrac{18\cdot22}{\left(20-1\right)\left(20+1\right)}\)
\(=\dfrac{1\cdot5}{2\cdot4}\cdot\dfrac{2\cdot6}{3\cdot5}\cdot...\cdot\dfrac{18\cdot22}{19\cdot21}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot21\cdot22}{2\cdot3\cdot4\cdot5\cdot...\cdot19\cdot20\cdot21}=1\cdot22=22\)
TH
16 tháng 11 2017
\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)
=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)
Từ giả thiết \(\Rightarrow\left\{{}\begin{matrix}x\inƯ\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\\x+1\inƯ\left(15\right)=\left\{1;3;5;15\right\}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{1;2;3;4;6;8;12;24\right\}\\x\in\left\{0;2;4;14\right\}\end{matrix}\right.\)
Vậy \(x\in\left\{2;4\right\}\)