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 = \(\varnothing\)
3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)
\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)
\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)
\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)
\(\left|\dfrac{3}{2}-x\right|=-2\)
=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )
Vậy x=\(\varnothing\)
4) \(\left(-\dfrac{4}{3}\right)-\left|2x+\left(-1\right)\right|=-\dfrac{7}{3}\)
\(\left|2x+\left(-1\right)\right|=\left(-\dfrac{4}{3}\right)-\left(-\dfrac{7}{3}\right)\)
\(\left|2x+\left(-1\right)\right|=\left(-\dfrac{4}{3}\right)+\dfrac{7}{3}\)
\(\left|2x+\left(-1\right)\right|=1\)
=> \(2x+\left(-1\right)=\pm1\)
+)\(TH1:2x+\left(-1\right)=1\) +)\(TH2:2x+\left(-1\right)=-1\)
\(2x=1-\left(-1\right)\) \(2x=\left(-1\right)-\left(-1\right)\)
\(2x=1+1\) \(2x=\left(-1\right)+1\)
\(2x=2\) \(2x=0\)
\(x=2:2\) \(x=0:2\)
\(x=1\) \(x=0\)
Vậy x=1 Vậy x=0
5) \(\left(-9\right).\left|36-\dfrac{x}{2}\right|=\left(-54\right)\)
\(\left|36-\dfrac{x}{2}\right|=\left(-54\right):\left(-9\right)\)
\(\left|36-\dfrac{x}{2}\right|=6\)
=> \(36-\dfrac{x}{2}=\pm6\)
+)\(TH1:36-\dfrac{x}{2}=6\) +)\(TH2:36-\dfrac{x}{2}=-6\)
\(\dfrac{x}{2}=36-6\) \(\dfrac{x}{2}=36-\left(-6\right)\)
\(\dfrac{x}{2}=30\) \(\dfrac{x}{2}=42\)
\(x=30.2\) \(x=42.2\)
\(x=60\) \(x=84\)
Vậy x=60 Vậy x=84
6) \(0,75-\left|\left(-x\right)+\dfrac{1}{3}\right|=0,25\)
\(\dfrac{3}{4}-\left|\left(-x\right)+\dfrac{1}{3}\right|=\dfrac{1}{4}\)
\(\left|\left(-x\right)+\dfrac{1}{3}\right|=\dfrac{3}{4}-\dfrac{1}{4}\)
\(\left|\left(-x\right)+\dfrac{1}{3}\right|=\dfrac{2}{4}\)
\(\left|\left(-x\right)+\dfrac{1}{3}\right|=\dfrac{1}{2}\)
Vì giá trị tuyệt đối luôn thuộc N , mà \(\dfrac{1}{2}\) lại là phân số nên \(\left(-x\right)+\dfrac{1}{3}\) không có giá trị của x
Vậy x=\(\varnothing\)
7)\(\left(-\dfrac{6}{5}\right)+\left|x-9\right|=-\dfrac{6}{5}\)
\(\left|x-9\right|=\left(-\dfrac{6}{5}\right)-\left(-\dfrac{6}{5}\right)\)
\(\left|x-9\right|=\left(-\dfrac{6}{5}\right)+\dfrac{6}{5}\)
\(\left|x-9\right|=1\)
=> \(x-9=\pm1\)
+)\(TH1:x-9=1\) +)\(TH2:x-9=-1\)
\(x=1+9\) \(x=\left(-1\right)+9\)
\(x=10\) \(x=8\)
Vậy x=10 Vậy x=8
8) \(\left(-4\right).\left|3x-\dfrac{1}{5}\right|=12\)
\(\left|3x-\dfrac{1}{5}\right|=12:\left(-4\right)\)
\(\left|3x-\dfrac{1}{5}\right|=-3\)
Vì giá trị tuyệt đối luôn luôn lớn hơn 0
=> Không có giá trị nào của x thỏa mãn
Vậy x=\(\varnothing\)