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a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)
-4 <x<\(\frac{-3}{10}\)
\(\frac{-40}{10}\)< x <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}
b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)
\(\frac{-875}{189}< x< \frac{108}{189}\)
=> x E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Bài 1: Tìm x biết:
1) x +\(\frac{7}{12}\)= \(\frac{17}{18}\)- \(\frac{1}{9}\) 2) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+ x) = \(\frac{7}{69}\)
x +\(\frac{7}{12}\)= \(\frac{15}{18}\) \(\frac{13}{23}\)+ x = \(\frac{29}{30}\)- \(\frac{7}{69}\)
x = \(\frac{15}{18}\)- \(\frac{7}{12}\) \(\frac{13}{23}\)+ x = \(\frac{199}{230}\)
x = \(\frac{1}{4}\) x = \(\frac{3}{10}\)
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
a) ta có \(\frac{-5}{6}\)\(\times\)\(\frac{120}{25}\)< \(x\)<\(\frac{-7}{15}\)\(\times\)\(\frac{4}{9}\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2074074074\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)( -1;-2;-3)
b) ta có \(\left(\frac{-5}{3}\right)^3\)<\(x\)<\(\frac{-25}{35}\)\(\times\)\(\frac{-5}{6}\)\(\Rightarrow\)\(-4,62962963\)<\(x\)<\(0,5952380952\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)(-4;-3;-2;-1;0)
ĐÚNG THÌ K CHO MK NHA
\(a)\frac{8}{9}x-\frac{2}{3}=\frac{1}{3}x+1\frac{1}{3}\)
\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+1\frac{1}{3}\)
\(\Rightarrow\frac{5}{9}x=\frac{2}{3}+\frac{4}{3}\)
\(\Rightarrow\frac{5}{9}x=2\Rightarrow x=2\div\frac{5}{9}=\frac{18}{5}\)
\(b)(\frac{-2}{5}+\frac{3}{7})-(\frac{4}{9}+\frac{12}{20}-\frac{13}{25})+\frac{7}{35}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{3}{5}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{15}{25}-\frac{13}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-(\frac{4}{9}+\frac{2}{25})+\frac{1}{5}\)
\(=\frac{1}{35}-\frac{118}{25}+\frac{1}{5}\)
Làm nốt
1) \(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\Leftrightarrow\frac{-5}{6}.\frac{24}{5}< x< \frac{-63}{210}\)
\(\Leftrightarrow-40< x< \frac{-63}{210}\)
\(\Leftrightarrow\frac{-400}{10}< \frac{10x}{10}< \frac{-3}{10}\)
\(\Leftrightarrow-400< 10x< -3\)
\(\Leftrightarrow x\in\left\{-39;-38;...;-2;-1\right\}\)
2) \(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)
\(\Leftrightarrow\frac{-125}{25}< x< \frac{4}{7}\)
\(\Leftrightarrow\frac{-35}{7}< \frac{-7x}{7}< \frac{4}{7}\)
\(\Leftrightarrow-35< -7x< 4\)
\(\Leftrightarrow x\in\left\{4;3;2;1;0\right\}\)