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\(\frac{2\frac{1}{2}x-1}{\frac{2}{3}}=\frac{\frac{-2}{3}}{1-2\frac{1}{2}x}\) ĐKXĐ \(x\ne\frac{2}{5}\)
\(\Leftrightarrow\)\(\frac{\frac{5}{2}x-1}{\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{5}{2}x-1}\)\(\Leftrightarrow\)\(\left(\frac{5}{2}x-1\right)^2=\frac{4}{9}\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+1=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+\frac{5}{9}=0\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-\frac{25}{6}x-\frac{5}{6}x+\frac{5}{9}=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{4}x^2-\frac{25}{6}x\right)-\left(\frac{5}{6}x-\frac{5}{9}\right)=0\)\(\Leftrightarrow\)\(\frac{25}{2}x\left(\frac{1}{2}x-\frac{1}{3}\right)-\frac{5}{3}\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{2}x-\frac{5}{3}\right)\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{2}{15}\end{cases}}\)
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+6y-1}{5x}\left(1\right)\)
Từ `2` tỉ số đầu , ta áp dụng t/c của DTSBN , ta đc :
\(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=\dfrac{2x+3+3y-2}{3+6}=\dfrac{2x+3y+1}{9}\left(2\right)\)
Từ `(1);(2)=>`\(\dfrac{2x+6y-1}{5x}=\dfrac{2x+3y+1}{9}\left(3\right)\)
Từ `(3)` ta xét `2` trường hợp :
+, Nếu `2x+3y+1 \ne 0` thì :
`(3)=>5x=9=>x=9/5`
Thay `x=9/5` vào \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}\), ta đc :
\(\dfrac{2\cdot\dfrac{9}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{\dfrac{18}{5}+3}{3}=\dfrac{3y-2}{6}\\ \Rightarrow\dfrac{11}{5}=\dfrac{3y-2}{6}\\ 3y-2=6\cdot\dfrac{11}{5}\\ 3y-2=\dfrac{66}{5}\\ 3y=\dfrac{76}{5}\\ y=\dfrac{76}{16}\)
+, Nếu `2x+3y+1=0` thì :
`(1)=>` \(\dfrac{2x+3}{3}=\dfrac{3y-2}{6}=0\\ \Rightarrow\left\{{}\begin{matrix}2x+3=0\\3y-2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)
4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
Tích chéo ta có:
\(-(2\frac{1}{2}x-1) ^2=-(\frac{2}{3})^2 \)
<=>\(2\frac{1}{2}x -1=\frac{2}{3} \)
<=>\(2\frac{1}{2}x =\frac{5}{3} \)
<=>\(\frac{5}{2}x=\frac{5}{3} \)
<=>\(x=\frac{5}{3}:\frac{5}{2} \)
<=>\(x=\frac{2}{3} \)
bạn giải thích rõ chỗ :\(\left(2\dfrac{1}{2}x-1\right)\times\left(1-2\dfrac{1}{2}x\right)=-\left(2\dfrac{1}{2}x-1\right)^2\)hộ mik với