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5/9 - 2/3 = -15/9x + 1 2/9x
-1/9 = (-15/9 + 1 2/9)x
-1/9 = -4/9x
x = -1/9 :-4/9
x = 1/4
vậy x = 1/4
đúng 100% đấy bạn ơi
\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)
\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)
\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)
\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)
\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)=x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{2016}>\frac{1}{2017}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}>\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\)
=> x = 0
Vậy x = 0
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)=0\)
\(\Rightarrow x=0\).Do \(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\ne0\)
Vậy x=0
a) Ta có:
\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-x-\frac{1}{4}\\ \Rightarrow x+\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-\frac{1}{4}\\ \Rightarrow x>\frac{2}{3}+\frac{4}{9}-\frac{1}{4}-\frac{1}{6}-\frac{4}{15}\\ \Rightarrow x>\left(\frac{6}{9}+\frac{4}{9}\right)-\left(\frac{15}{60}+\frac{10}{60}+\frac{16}{60}\right)\)
\(x>\frac{10}{9}-\frac{41}{60}\\ x>\frac{200-123}{180}\Rightarrow x>\frac{77}{180}\)
b) Bất đẳng thức kép
\(4-1\frac{1}{3}< x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
có nghĩa là ta phải có hai bất đẳng thức đồng thời:
\(x+\frac{1}{5}>4-1\frac{1}{3}\) và \(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
Ta tìm các giá trị của x cần thỏa mãn bất đẳng thức thứ nhất:
\(x+\frac{1}{5}>4-1\frac{1}{3}\Rightarrow x>4-1\frac{1}{3}-\frac{1}{5}\\ \Rightarrow x>\frac{37}{15}\)
Từ bất đẳng thức thứ hai
\(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\Rightarrow x< \frac{86}{7}-\frac{27}{8}-\frac{1}{5}\\ \Rightarrow x< \frac{2439}{280}.\)
Như vậy các số hữu tỉ x cần thỏa mãn:
\(\frac{37}{15}< x< \frac{2439}{280}\)
\(\frac{5}{9}-1^2_9x=\frac{2}{3}-\frac{15}{9}z\)
\(\frac{15}{9}x-\frac{11}{9}x=\frac{2}{3}-\frac{5}{9}\)
\(\frac{4}{9}x=-\frac{1}{9}\)
\(x=-\frac{1}{4}\)