1 - (\(11\frac{1}{2}\)- 10,1 + x ) : \(8\frac{2}{5}\)= 0
Tìm X
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(1/1×2 + 1/2×3 + ... + 1/9×10) × x < 2/1×3 + 2/3×5 + ... + 2/9×11
(1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10) × x < 1 - 1/3 + 1/3 - 1/5 + ... + 1/9 - 1/11
(1 - 1/10) × x < 1 - 1/11
9/10 × x < 10/11
x < 10/11 : 9/10
x < 10/11 × 10/9
x < 100/99
Mà x là số tự nhiên => x = 0 hoặc 1
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
\(\frac{2}{3}x-\frac{3}{2}\left(x-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}x-\frac{3}{2}x+\frac{3}{4}=\frac{5}{12}\)
\(\Rightarrow\frac{-5}{6}x=\frac{5}{12}-\frac{3}{4}=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{3}:\frac{-5}{6}=\frac{2}{5}\)
vậy x = \(\frac{2}{5}\)
= 2/3x - 3/2x - 3/2 × 1/2= 5/12
=> -5/6x - 3/4 = 5/12
=> -5/6x = 5/12 +3/4= 7/6
=>x=7/6 ÷ -5/6 =1/3
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\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
\(\left(\frac{5}{3}-\frac{2}{5}\right)x=\frac{19}{10}\)
\(\frac{19}{15}x=\frac{19}{10}\)
\(x=\frac{19}{30}\)
\(\frac{5}{3}x-\frac{2}{5}x=\frac{19}{10}\)
(5/3 - 2/5)x = 19/10
19/15x = 19/10
x = 19/30
b) \(\frac{4}{9}x-\frac{1}{2}=\frac{-5}{9}\)
\(\Rightarrow\frac{4}{9}x=\frac{-5}{9}+\frac{1}{2}\)
\(\Rightarrow\frac{4}{9}x=\frac{-1}{18}\)
\(\Rightarrow x=\frac{-1}{18}:\frac{4}{9}\)
\(\Rightarrow x=\frac{-1}{8}\)
\(\frac{2}{7}< \frac{x}{3}< \frac{11}{4};x\inℕ\)
=>\(\frac{12.2}{84}< \frac{28x}{84}< \frac{11.21}{84}\)
=>\(\frac{24}{84}< \frac{28x}{84}< \frac{231}{84}\)
=>24<28x<231
=>28x\(\in\){25;26;27;28;.............................;230}
=>Các số chia hết cho 28 là:28;56;84;112;140;168;196;224
=>x (thỏa mãn)\(\in\){1;2;3;4;5;6;7;8}
Vậy x\(\in\) {1;2;3;4;5;6;7;8}
\(\left(4,5m-\frac{3}{4}.5\frac{1}{3}\right).\frac{1}{12}+\frac{1}{2}x=1\frac{1}{2}\)
\(\left(4,5m-\frac{3}{4}.\frac{16}{3}\right).\frac{1}{2}.\frac{1}{6}+\frac{1}{2}x=\frac{3}{2}\)
\(\left(4,5m-\frac{48}{12}\right).\frac{1}{2}.\left(\frac{1}{6}+x\right)=\frac{3}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}:\frac{1}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{3}{2}.\frac{2}{1}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=\frac{6}{2}\)
\(\left(4,5m-4\right).\left(\frac{1}{6}+x\right)=3\)
=>3\(⋮\)\(\frac{1}{6}+x\)
=>\(\frac{1}{6}+x\)\(\in\)Ư(3)={\(\pm\)1;\(\pm\)3}
Ta có bảng:
\(\frac{1}{6}+x\) | -1 | 1 | -3 | 3 |
x | \(-1\frac{1}{6}\) | \(1\frac{1}{6}\) | \(-3\frac{1}{6}\) | 3\(\frac{1}{6}\) |
Vậy x\(\in\){\(-1\frac{1}{6}\);\(1\frac{1}{6}\);\(-3\frac{1}{6}\);\(\frac{1}{6}\)}
Chúc bn học tốt
\(1-\left(11\frac{1}{2}-10,1+x\right):8\frac{2}{5}=0\)
=>\(1-\left(\frac{21}{2}-\frac{101}{10}+x\right):\frac{42}{5}=0\)
=> \(1-\left(\frac{2}{5}+x\right)=0\)
=>\(1=\frac{2}{5}+x\)
=>\(x=1-\frac{2}{5}\)
=>\(x=\frac{3}{5}\)
Vậy ..................