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Ta có : \(x-\frac{3}{1.5}-\frac{3}{5.9}-\frac{3}{9.13}-\frac{3}{13.17}-\frac{3}{17.21}=\frac{2}{7}\)
=> \(x-\left(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+\frac{3}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{3}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}.\frac{20}{21}=\frac{2}{7}\)
=> \(x-\frac{5}{7}=\frac{2}{7}\)
=> x = 1
Vậy x = 1
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000000000000000000000000000000000000000000000000000000000000000000000000000000ololllllllllllololol
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a) -x+\(\frac{2}{5}\)=\(\frac{3}{10}\)
-x=\(\frac{3}{10}\)\(-\)\(\frac{2}{5}\)
\(-1.x\)=\(\frac{3}{10}-\frac{4}{10}\)
\(-1.x=\frac{-1}{10}\)
\(x=\frac{-1}{10}:-1\)
\(x=\frac{1}{10}\)
b)\(-x+\frac{3}{10}=1\)
\(-x=1-\frac{3}{10}\)
\(-x=\frac{10}{10}-\frac{3}{10}\)
\(-x=\frac{7}{10}\)
\(=>x=\frac{-7}{10}\)
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\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)
\(=\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)
\(=\frac{59}{15}-\frac{35}{6}:\frac{7}{4}\)
\(=\frac{59}{15}-\frac{10}{3}\)
\(=\frac{9}{15}=\frac{3}{5}\)
\(\cdot62,87+35,14+4,13+8,35+4,86+5,65\)
\(=\left(62,87+4,13\right)+\left(35,14+4,86\right)+\left(8,35+5,65\right)\)
\(=67+40+14\)
\(=121\)
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\(2|3x-1|=4+|3x-1|\)
\(2|3x-1|-|3x-1|=4\)
\(|3x-1|=4\)
\(\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}}\)
\(\orbr{\begin{cases}3x=4+1\\3x=-4+1\end{cases}}\)
\(\orbr{\begin{cases}3x=5\\3x=-3\end{cases}}\)
\(\orbr{\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
Vậy x =\(\frac{5}{3}\)hoặc x=\(-1\)
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Mik có nhầm lẫn nhé
l2x - 1l = 16
suy ra 2x-1 = 16 và -16
nếu 2x-1 = 16 nếu 2x-1 = -16
2x = 16+1 2x = (-16) +1
2x=17 2x = -15
x=17:2 x = (-15) :2
x = 8,5 x = -7,5
Vậy x = 8,5 và -7,5
Vì ko có giá trị tuyệt đối nào có giá trị bằng 0
suy ra : x ko có giá trị
\(9^{x+1}=27^{x-2}\)
\(3^{2.\left(x+1\right)}=3^{3.\left(x-2\right)}\)
\(\Rightarrow2\left(x+1\right)=3.\left(x-2\right)\)
\(\Rightarrow2x+2=3x-6\)
\(\Rightarrow2x-3x=-6-2\)
\(\Rightarrow-x=-8\)
\(\Rightarrow x=8\)