BÀI TẬP: tìm X , biết :
a. \(\frac{1}{3}\)x + \(\frac{2}{5}\)( x - 1 ) = 0 b. (2x-3).(56-7x)=0
c.2./ \(\frac{1}{2}\)x -\(\frac{1}{3}\)/ - \(\frac{5}{8}\)=\(\frac{1}{4}\) d. (0,6x-\(\frac{1}{2}\) ) .\(\frac{3}{4}\)- (-1)=\(\frac{1}{3}\)
e.\(\frac{60}{100}\)x +\(\frac{2}{3}\)x=\(\frac{1}{3}\).\(6\frac{1}{3}\) f.\(\frac{2}{3}\)x-\(\frac{50}{100}\)x=\(\frac{5}{12}\)
g.\(\left(3\frac{1}{2}+2x\right)\)\(\div\)\(\frac{3}{14}\)=\(\frac{7}{12}\) h.\(\left(2\frac{1}{4}-1\frac{4}{5}\right)\)x-\(\frac{3}{20}\)=1
i.\(\left(x+0,2\right)^2\) +\(\frac{17}{25}\) =\(1\frac{1}{25}\) j.\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)
k.\(\frac{x-3}{x-5}=\left(\frac{-3}{5}\right)^2\) l.\(\frac{x-2}{3}\) \(=\frac{3-2x}{2^2}\)
a) \(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\) b) ( 2x - 3 ) . ( 56 - 7x ) = 0
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\) \(\Rightarrow\orbr{\begin{cases}2x-3=0\\56-7x=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\7x=56\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=8\end{cases}}\)
\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)
\(x.\frac{11}{15}-\frac{2}{5}=0\)
\(x.\frac{11}{15}=0+\frac{2}{5}=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{11}{15}=\frac{6}{11}\)
c) \(2.\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{5}{8}=\frac{1}{4}\)
\(2.\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{5}{8}\)
\(2.\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}:2\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{16}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{16}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{16}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{7}{16}+\frac{1}{3}=\frac{37}{48}\\\frac{1}{2}x=\frac{-7}{16}+\frac{1}{3}=\frac{-5}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{37}{24}\\x=\frac{-5}{24}\end{cases}}\)
d) \(\left(0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\left(0,6x-\frac{1}{2}\right).\frac{3}{4}+1=\frac{1}{3}\)
\(\left(0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{1}{3}-1\)
\(\left(0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{-2}{3}\)
\(\left(0,6x-\frac{1}{2}\right)=\frac{-2}{3}:\frac{3}{4}\)
\(0,6x-\frac{1}{2}=\frac{-8}{9}\)
\(0,6x=\frac{-8}{9}+\frac{1}{2}\)
\(0,6x=-\frac{7}{18}\)
\(x=\frac{-7}{18}:0,6\)
\(x=\frac{-35}{54}\)
a) <=> \(\frac{11}{15}x=\frac{2}{5}\)<=> \(x=\frac{6}{11}\)
b)<=>\(\orbr{\begin{cases}2x-3=0\\56-7x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{3}{2}\\x=8\end{cases}}\)
c) <=> \(2:\frac{x}{2}-\frac{1}{3}:\left(-\frac{5}{8}\right)=\frac{1}{4}\)
<=>\(\frac{4}{x}-\left(-\frac{8}{15}\right)=\frac{1}{4}\)
<=> 4/x = -17/60
<=> x= -240/17
d) <=>\(\left(\frac{3}{5}x-\frac{1}{2}\right).\frac{3}{4}=\frac{-2}{3}\)
<=>\(\frac{9}{20}x=-\frac{7}{24}\)
<=>\(x=\frac{-35}{54}\)
e) <=> \(\frac{19}{15}x=\frac{19}{9}\)
<=> \(x=\frac{5}{3}\)
f) <=> \(\frac{1}{6}x=\frac{5}{12}\)
<=> \(x=\frac{5}{2}\)
g) <=> \(\frac{49}{3}+\frac{28}{3}x=\frac{7}{12}\)
<=>\(\frac{28}{3}x=-\frac{63}{4}\)
<=> x= -27/16
h) <=>\(\frac{9}{20}x=\frac{23}{20}\)
<=> x=23/9
i) (x+0,2)2 =9/25
<=> \(\orbr{\begin{cases}x+0,2=\frac{3}{5}\\x+0,2=-\frac{3}{5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
j) <=> \(-\left(3x-\frac{7}{9}\right)^3=\frac{8}{27}\)
<=> \(3x-\frac{7}{9}=\frac{2}{3}\)<=> x= 13/27
k) <=> \(\frac{x-3}{x-5}=\frac{9}{25}\)
<=> (x-3).25 = 9.(x-5)
<=>25x-9x=75-45
<=> x=15/8
l) <=> (x-2).22 = (3-2x).3
<=> 4x - 8 = 9 - 6x
<=> 10x = 17
<=> x=1,7
