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mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
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\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)
TH1 : \(\Rightarrow x+\frac{1}{2}=0\)
\(x=0-\frac{1}{2}\)
\(x=\frac{-1}{2}\)
TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)
\(2x=0+\frac{2}{3}=\frac{2}{3}\)
\(x=\frac{2}{3}\div2=\frac{1}{3}\)
\(\Rightarrow x=\frac{-1}{2};\frac{1}{3}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
\(x=\frac{3}{5}^2-\frac{1}{5}^2\)
\(x=\frac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x:2+x:\frac{5}{2}+x:\frac{1}{3}=\frac{-18}{25}\)
\(x.\frac{1}{2}+x.\frac{2}{5}+x.3=\frac{-18}{25}\)
\(x.\left(\frac{1}{2}+\frac{2}{5}+3\right)=\frac{-18}{25}\)
\(x.\frac{39}{10}=\frac{-18}{25}\)
\(x=\frac{-18}{25}:\frac{39}{10}\)
\(x=\frac{-12}{65}\)
Chúc bn học tốt !!!
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\(a,\frac{5}{7}+\frac{2}{7}:x=\frac{-11}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-11}{14}-\frac{10}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-21}{14}\)
\(\Rightarrow\frac{2}{7}:x=\frac{-3}{2}\)
Tự tìm x được rồi :v
\(b,\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Rightarrow\left[\frac{2}{3}-\frac{1}{2}\right]x=\frac{5}{12}\)
\(\Rightarrow\left[\frac{4}{6}-\frac{3}{6}\right]x=\frac{5}{12}\)
\(\Rightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\frac{1}{6}=\frac{5}{12}\cdot\frac{6}{1}=\frac{5}{2}\cdot\frac{1}{1}=\frac{5}{2}\)
\(c,\left[x+\frac{1}{2}\right]^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left[x+\frac{1}{2}\right]^2=\frac{26}{25}-\frac{17}{25}\)
\(\Rightarrow\left[x+\frac{1}{2}\right]^2=\frac{9}{25}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{3}{5}\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=\frac{3}{5}\\x+\frac{1}{2}=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{10}\\x=-\frac{11}{10}\end{cases}}\)
d, Tự làm
P/S : Thông cảm cho mk nhé vì máy tính mk ko bấm dấu ngoặc tròn được :v
\(-3\frac{3}{5}.x=\frac{4}{15}\)
\(\frac{-18}{5}.x=\frac{4}{15}\)
\(x=\frac{4}{15}:\frac{-18}{5}\)
\(x=\frac{-2}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
\(\frac{3}{2}-2x=25\%\)
\(\frac{3}{2}-2x=\frac{1}{4}\)
\(2x=\frac{3}{2}-\frac{1}{4}\)
\(2x=\frac{5}{4}\)
\(x=\frac{5}{4}:2\)
\(x=\frac{5}{8}\)
\(\frac{3}{2}\)- 2.x = 25%
\(\frac{3}{2}\)- 2.x = \(\frac{1}{4}\)
2 . x = \(\frac{3}{2}-\frac{1}{4}\)
2 . x = \(\frac{5}{4}\)
x = \(\frac{5}{4}\): 2
x = \(\frac{5}{8}\)