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TN
30 tháng 12 2022
b) \(\left(x-1\right)^3=\dfrac{1}{8}\)
\(\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\)
\(x-1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+1\)
\(x=\dfrac{3}{2}\)
3 tháng 8 2016
1)\(x+0,5+x+1,5+x+2,5=33\)
\(\Leftrightarrow3x=33-0,5-1,5-2,5=28,5\)
\(\Leftrightarrow x=9,5\)
2)\(\left(x+0,9\right)\left(1-0,4\right)=2412\)
\(\Leftrightarrow\left(x+0,9\right)\cdot0,6=2412\)
\(\Leftrightarrow x+0,9=4020\)
\(\Leftrightarrow x=1019,1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 5 2023
(5 - \(x\))(9\(x^2\) - 4) =0
\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 5 2023
72\(x\) + 72\(x\) + 3 = 344
72\(x\) \(\times\) ( 1 + 73) = 344
72\(x\) \(\times\) (1 + 343) = 344
72\(x\) \(\times\) 344 = 344
72\(x\) = 344 : 344
72\(x\) = 1
72\(x\) = 70
\(2x\) = 0
\(x\) = 0
Kết luận: \(x\) = 0
14 tháng 3 2023
\(\dfrac{0.375-0.3+\dfrac{3}{11}+\dfrac{3}{12}}{-0.625+0.5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1.5+1-0.75}{2.5+\dfrac{5}{3}-1.25}\)
=\(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
=\(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
=\(\dfrac{3}{-5}+\dfrac{3}{5}\)
=\(0\)
NK
6 tháng 7 2023
\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)
\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)
\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)
\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)
\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)
\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)
\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)
\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)
\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)
\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)
\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)
\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)
\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)
\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)
\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)
\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)
\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)
\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)
\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)
\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)
\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)
30 tháng 9 2016
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{3004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{3}{15}-\frac{10}{15}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
27 tháng 6 2018
Bài 1:
ta có: 333<3333; 444<4444
=> 333444<33334444
Bài 2:
\(A=\frac{21^5}{81}=\frac{\left(3.7\right)^5}{3^4}=\frac{3^5.7^5}{3^4}=3.7^5=50421\)
\(B=\frac{3^3.\left(0,5\right)^5}{\left(1,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{\left(3.0,5\right)^5}=\frac{3^3.\left(0,5\right)^5}{3^5.\left(0,5\right)^5}=\frac{1}{3^2}=\frac{1}{9}\)
\(C=2^2.\frac{1}{128}.45.2^{-6}=\frac{2^2.45}{128.64}=\frac{2^2.45}{2^7.2^6}=\frac{45}{2^{11}}=\frac{45}{2048}\)
\(D=\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+2^2.3^3+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}\)\(=3^3.\left(-1\right)=-27\)
6 tháng 8 2017
\(\left(\frac{1}{4}x-1,5\right)+\left(\frac{5}{6}-3\right)-\left(\frac{5}{8}x-0,5\right)=45\)
\(\frac{1}{4}x-1,5-\frac{13}{6}-\frac{5}{8}x+0,5=45\)
\(-\frac{3}{8}x-\frac{19}{6}=45\)
\(-\frac{3}{8}x=\frac{289}{6}\)
\(\Rightarrow x=-\frac{1156}{9}\)
6 tháng 8 2017
cảm ơn bạn anh nha mà cho em hỏi tại sao x lại ghép đc với nhau z?
\(0,5^3+1^3+1,5^3+...+5^3\)
\(=0,5^3\left(1^3+2^3+3^3+...+10^3\right)\)
\(=0,5^3.\left[\frac{10\left(10+1\right)}{2}\right]^2=378,125\)