tìm x biết : \(\frac{x}{27}\)-\(\frac{2}{9}\)=\(\frac{-1}{3}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
\(\frac{x-1}{27}=\frac{-3}{1-x}\)
\(\Rightarrow\frac{x-1}{27}=\frac{3}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=81\)
\(\Rightarrow x-1=\pm9\)
Làm nốt nhassssso
\(a,\frac{27}{x+1}=\frac{x+1}{3}\)
\(\Leftrightarrow27.3=\left(x+1\right)\left(x+1\right)\)
\(\Leftrightarrow81=\left(x+1\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=9^2\\\left(x+1\right)^2=\left(-9\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}}\)
Vậy \(x\in\left\{8;-10\right\}\)
a)\(\frac{27}{x+1}=\frac{x+1}{3}\)
Ta có : 27 . 3 = (x + 1)(x + 1)
=> 81 = (x + 1)2
=> 92 = (x + 1)2
=> \(\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}\)
\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)
\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)
\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)
\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)
\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)
bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.