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a) ta co:
1/18<x/12<y/9<1/4
=>2/36<x.3/36<y.4/36<9/36
=>x.3thuộc{3;6};y.4thuộc{4;8}
=>x thuộc{1;2};y thuộc{1:2}
b) ta co
7/8<x/40<9/10
=>70/80<x.2/40<72/80
=>x.2 =71
=>x=71/2
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
=5/12*2/3+2/3*7/12+1/3
=2/3*(5/12+7/12)+1/3
=2/3*12/12+1/3
=2/3*1+1/3
=2/3+1/3
=3/3
=1
đây là cách làm chi tiết bạn có thể bỏ bớt đc 1 số chỗ
M = 1/21 + 1/28+1/36+...+1/465
= 2/42+2/56+2/72+...+2/930
= 2.( 1/6.7 + 1/7.8 + 1/ 7.9 + ... + 1/30.31)
= 2.( 1/6-1/7+1/7-1/8+...+1/30-1/31)
= 2.(1/6 - 1/31) = 2.25/186 = 25/92
<=> |x+2| = 13
<=> \(\orbr{\begin{cases}x+2=13\\x+2=-13\end{cases}\Rightarrow}\orbr{\begin{cases}x=11\\x=-15\end{cases}}\)
Vậy.........
hok tốt
..........
\(|x+2|=12+\left(-3\right)+\left|-4\right|\)
\(|x+2|=12-3+4\)
\(\left|x+2\right|=13\)
\(\Rightarrow x\in\left\{-15;11\right\}\)
tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi
A=1/2+1/6+1/12+...+1/9900
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100
=99/100
\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251
\(12.\left(\dfrac{29292929}{56565656}+\dfrac{66666}{77777}\right)=12.\left(\dfrac{29}{56}+\dfrac{6}{7}\right)=12.\left(\dfrac{29}{56}+\dfrac{48}{56}\right)=12.\dfrac{77}{56}=\dfrac{33}{2}\)
\(12.\left(\dfrac{29292929}{56565656}+\dfrac{66666}{77777}\right)=12.\left(\dfrac{29.1010101}{56.1010101}+\dfrac{6.11111}{7.11111}\right)=12.\left(\dfrac{29}{56}+\dfrac{6}{7}\right)=12.\dfrac{11}{8}=\dfrac{33}{2}\)