3/8 x (X - 2/3) - 2/9 x (3X - 1/2) = 7/36
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b)(x+3)2-(x-4)(x+8)=1
\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1
⇒x2+6x+9-x2-8x+4x+32=1
⇒2x+41=1
\(\Rightarrow\)2x+41-1=0
\(\Rightarrow\)2x+40=0
⇒2x=-40
\(\Rightarrow\)x=\(\dfrac{-40}{2}\)
⇒x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{8}\left(x-\dfrac{2}{3}\right)-\dfrac{2}{9}\left(3x-\dfrac{1}{2}\right)=-\dfrac{7}{36}\)
\(\Leftrightarrow\dfrac{3}{8}x-\dfrac{1}{4}-\dfrac{2}{3}x+\dfrac{1}{9}=-\dfrac{7}{36}\)
\(\Leftrightarrow\dfrac{-7}{24}x=-\dfrac{1}{18}\)\(\Leftrightarrow x=\dfrac{4}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
\(\dfrac{3}{8}x-\dfrac{1}{4}-\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{7}{36}\)
=> \(\dfrac{3}{8}x-\dfrac{2}{3}x-\left(\dfrac{1}{4}+\dfrac{1}{9}\right)=\dfrac{7}{36}\)
=> \(\dfrac{-7}{24}x-\dfrac{5}{36}=\dfrac{7}{36}\)
=> \(\dfrac{-7}{24}x=\dfrac{1}{3}\)
=> \(x=\dfrac{-8}{7}\)