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\(\left|4x\right|-\left|-13,5\right|=\left|-7,5\right|\)
\(\Rightarrow\left|4x\right|-13,5=7,5\)
\(\Rightarrow\left|4x\right|=21\)
\(\Rightarrow\left[{}\begin{matrix}4x=21\\4x=-21\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{21}{4}\\x=\dfrac{-21}{4}\end{matrix}\right.\)
\(\left|x-3,4\right|+\left|2,6-y\right|=0\)
\(\left\{{}\begin{matrix}\left|x-3,4\right|\ge0\forall x\\\left|2,6-y\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-3,4\right|+\left|2,6-y\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-3,4\right|=0\Rightarrow x=3,4\\\left|2,6-y\right|=0\Rightarrow y=2,6\end{matrix}\right.\)
\(A= \left|x-500\right|+ \left|x-300\right|\)
\(A= \left|x-500\right|+\left|300-x\right|\)
Áp dụng bất đẳng thức:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(A\ge\left|x-500+300-x\right|\)
\(A\ge200\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-500\ge0\Rightarrow x\ge500\\300-x\ge0\Rightarrow x\le300\end{matrix}\right.\\\left\{{}\begin{matrix}x-500< 0\Rightarrow x< 500\\300-x< 0\Rightarrow x>300\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow300< x< 500\)
B1: \(\left|4x\right|-\left|-13.5\right|=\left|-7,5\right|\)
\(\Leftrightarrow\left|4x\right|-13,5=7,5\)
\(\left|4x\right|=7,5+13,5\)
\(\Rightarrow\left|4x\right|=21\)
\(\Rightarrow x\pm21\)
\(\cdot4x=21\)
\(x=21:4\)
\(x=\dfrac{21}{4}\)
\(\cdot4x=-21\)
\(x=-21:4\)
\(x=\dfrac{-21}{4}\)
\(Vậy\) \(x\)\(\in\)\(\left\{\dfrac{21}{4};\dfrac{-21}{4}\right\}\)
B2:
\(\left|x-3,4\right|+\left|2,6-y\right|=0\)
\(\Rightarrow\)... tự làm nhé bạn
2. để Bmax thì x+2/3 đạt GTNN=> x+2/3=0=>x=-2/3
3. 4x=21
4x=-21 tự tính
x-1.5=2
x-1.5=-2
x+3/4=1/2
x+3/4=-1/2
Bài : 5
a) Ta có : A = 3 + |4 - x|
Vì : \(\left|4-x\right|\ge0\forall x\)
Nên : A = 3 + |4 - x| \(\ge3\forall x\)
Vậy Amin = 3 khi x = 4
b) Ta có : B = 5|1 - 4x| - 1
Vì \(\text{5|1 - 4x|}\ge0\forall x\)
Nên : B = 5|1 - 4x| - 1 \(\ge-1\forall x\)
Vậy Bmin = -1 khi x = 1/4
a)\(\left|2x-3\right|=6\)
\(\Rightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}...\\...\end{cases}}\)
b)\(2.\left|3x+1\right|=5\)
\(\left|3x+1\right|=2,5\)
\(\Rightarrow\orbr{\begin{cases}3x+1=2,5\\3x+1=-2,5\end{cases}}\Rightarrow\orbr{\begin{cases}...\\...\end{cases}}\)
c)\(7,5-3\left|5-2x\right|=-4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=4\)
\(...\)
a) \(2,5:4x=0,5:0,2\)
\(2,5:4x=\frac{5}{2}\)
\(4x=2,5:\frac{5}{2}\)
\(4x=1\)
\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
b) \(\frac{1}{5}.x:3=\frac{2}{3}:0,25\)
\(\frac{1}{5}.x:3=\frac{8}{3}\)
\(\frac{1}{5}.x=\frac{8}{3}.3\)
\(\frac{1}{5}.x=8\)
\(x=8:\frac{1}{5}\)
\(x=40\)
Vậy \(x=40\)
a) \(\frac{2,5}{4x}=\frac{0,5}{0,2}\)
\(=>4x=\frac{0,2.2,5}{0,5}=1\)
\(=>x=\frac{1}{4}\)
b) \(\frac{1}{5}.\frac{x}{3}=\frac{2}{3}:0,25\)
\(=>\frac{x}{15}=\frac{4}{3}\)
\(=>x=\frac{4.15}{3}=20\)
mở dấu trị tuyệt đối ra rồi tính như bình thường