1) Tìm x Biết
x-\(\frac{1}{3}\)=\(\frac{3}{4}\)
\(\frac{-1,25}{x}=\frac{1}{-8}\)
\(4\frac{1}{3}\div\frac{x}{4}=6:0,3\)
\((x-\frac{4}{9})^2=\frac{1}{4}\)
\(9^x\times81^x=3^{2010}\)
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\(x\times\frac{6}{25}=\frac{15}{-13}\)
x=\(\frac{15}{-13}\div\frac{6}{25}\)
x=\(-\frac{125}{26}\)
các câu còn lại làm tương tự nha!!!
\(1.x.\frac{6}{25}=\frac{15}{-13}\\ x=\frac{15}{-13}:\frac{6}{25}\\ x=-\frac{125}{26}\)
\(2.x:\frac{4}{10}=\frac{13}{-45}+\frac{8}{15}\\ x:\frac{4}{10}=\frac{11}{45}\\ x=\frac{11}{45}.\frac{4}{10}\\ x=\frac{22}{225}\)
\(3.\frac{3}{8}-\frac{1}{6}.x=\frac{1}{4}\\ \frac{1}{6}.x=\frac{3}{8}-\frac{1}{4}\\ \frac{1}{6}.x=\frac{1}{8}\\ x=\frac{1}{8}:\frac{1}{6}\\ x=\frac{3}{4}\)
\(4.\frac{1}{3}+\frac{1}{2}:x=-4\\ \frac{1}{2}:x=-4-\frac{1}{3}=-\frac{13}{3}\\ x=\frac{1}{2}:\left(-\frac{13}{3}\right)=-\frac{3}{26}\)
\(5.x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}=\frac{5}{6}\\ x=\frac{5}{6}-\frac{7}{12}\\ x=\frac{1}{4}\)
a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)
\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)
\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow x-2=\frac{-2}{3}\)
hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)
\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)
hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)
Vậy: \(x=\frac{13}{15}\)
c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)
\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)
\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)
\(\Leftrightarrow3x=-160\)
hay \(x=\frac{-160}{3}\)
Vậy: \(x=\frac{-160}{3}\)
d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)
\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)
a/ (x - 2)3 + \(\frac{8}{27}\) = 0
=> (x - 2)3 = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)
=> x - 2 = \(-\frac{2}{3}\)
=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)
b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)
=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)
=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)
=> \(x=\frac{13}{60}.4=\frac{13}{15}\)
c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)
=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)
=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)
=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)
=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
Làm đi mình k cho Nhók Me tại mình làm xong rồi chỉ muốn xem đúng không thôi
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:
\(\frac{37}{15}< x< \frac{2439}{280}\)
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
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