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b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
\(2\times\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(-10x+3x=\frac{4}{5}-\frac{3}{2}\)
\(-7x=-\frac{7}{10}\)
\(x=\frac{1}{10}\)
\(2\times\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Rightarrow\frac{4}{5}-3x+10x=\frac{3}{2}\)
\(\Rightarrow\frac{4}{5}-7x=\frac{3}{2}\)
\(\Rightarrow7x=\frac{4}{5}-\frac{3}{2}\)
\(\Rightarrow7x=\frac{-7}{10}\)
\(\Rightarrow x=\frac{\frac{-7}{10}}{7}\)
\(\Rightarrow x=\frac{-1}{10}\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=\frac{-2x}{3}+3x\left(\frac{15x+20-126}{90}\right)-\frac{5x^2-20x}{10}\)
\(M=\frac{-2x}{3}+3x.\frac{15x-106}{90}-\frac{5.\left(x^2-4x\right)}{10}\)
\(M=\frac{-2x}{3}+\frac{45x^2-318x}{90}-\frac{x^2-4x}{2}\)
1
2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x
=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)
=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)
=> \(x=\frac{1}{10}\)
2 .
\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
=>\(11x=\frac{1}{6}\)
=>x=\(\frac{1}{66}\)
3.
\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
=>\(-2x=\frac{-2}{3}\)
=>\(\frac{1}{3}\)
4. câu 4 ko hiểu bạn ơi
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
1,\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\frac{6}{4}-10x=\frac{4}{5}-3x\)
\(\frac{6}{4}+\frac{4}{5}=7x\)
\(\frac{23}{10}=7x\)
\(\frac{23}{70}=x\)
2,\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\frac{3}{2}-1-4x=\frac{2}{3}-7x\)
\(\frac{3}{2}-1-\frac{2}{3}=-3x\)
\(\frac{-1}{6}=-3x\)
\(\frac{1}{18}=x\)
3,\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
\(\frac{3}{2}-\frac{7}{6}+\frac{1}{3}=2x\)
\(\frac{2}{3}=2x\)
\(\frac{1}{3}=x\)
4,mình không hiểu a ở đây là gì
a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2
=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2
=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2
=> -4x + 6 = 0
=> -4x = -6
=> x = 3/2
b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)
\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=12\)
\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Leftrightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\frac{4}{5}-\frac{3}{2}\)
\(\Leftrightarrow-7x=-\frac{7}{10}\)
\(\Leftrightarrow x=\frac{1}{10}\)
\(\frac{6}{4}-10x=\frac{4}{5}-3x\)
\(3x-10x=\frac{4}{5}-\frac{6}{4}\)
\(-7x=-\frac{7}{10}\)
\(x=10\)
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