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Ta có: \(\hept{\begin{cases}\left|a\right|\ge0\\\left|b\right|\ge0\\\left|c\right|\ge0\end{cases}}\Rightarrow\left|a\right|+\left|b\right|+\left|c\right|\ge0\)
a)\(\Rightarrow\left|\frac{1}{4}-x\right|+\left|x-y+z\right|+\left|\frac{2}{3}+y\right|\ge0\)
\("="\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{2}{3}\\z=-\frac{11}{12}\end{cases}}\)
b) \(\Rightarrow\left|2-x\right|+\left|3-y\right|+\left|x+y+z\right|\ge0\)
\("="\Leftrightarrow\hept{\begin{cases}x=2\\y=3\\z=-5\end{cases}}\)
a) \(\left|\frac{1}{4}-x\right|+\left|x-y+z\right|+\left|\frac{2}{3}+y\right|=0\)
Ta có: \(\left|\frac{1}{4}-x\right|\ge0\)với mọi x
\(\left|x-y+z\right|\ge0\)vơi mọi x, y, z
\(\left|\frac{2}{3}+y\right|\ge0\) với mọi y
\(\left|\frac{1}{4}-x\right|+\left|x-y+z\right|+\left|\frac{2}{3}+y\right|\ge0\) với nọi x, y, z
Dấu "=" xảy ra khi và chỉ khi" \(\hept{\begin{cases}\frac{1}{4}-x=0\\x-y+z=0\\\frac{2}{3}+y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{2}{3}\\z=-\frac{11}{12}\end{cases}}\)
câu b cách làm giống như câu a
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
\(5^{\left(x-2\right).\left(x+3\right)}=1\)
mà \(5^0=1\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}\)
\(\left(x+1\right)^{200}=\left(2x-3\right)^{200}\)
\(\Rightarrow2x-x=1+3\)
\(\Rightarrow x=4\)
Vậy x = 4
\(\left(x+1\right)^{200}=\left(2x-3\right)^{200}\)
\(\Rightarrow x+1=2x-3\)
\(\Rightarrow x-2x=-3-1\)
\(\Rightarrow-x=-4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+3}\)
\(\Rightarrow\left(x+1\right)^{x+1}-\left(x+1\right)^{x+3}=0\)
\(\Rightarrow\left(x+1\right)^{x+1}.\left[1-\left(x+1\right)^2\right]=0\)
+) \(\left(x+1\right)^{x+1}=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
+) \(1-\left(x+1\right)^2=0\)
\(\Rightarrow\left(x+1\right)^2=1\)
\(\Rightarrow x+1=\pm1\)
+ \(x+1=1\Rightarrow x=0\)
+ \(x+1=-1\Rightarrow x=-2\)
Vậy \(x\in\left\{-2;-1;0\right\}\)
Ta có: \(\left|x-3\right|=\left|2x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2x=3+1\\x+2x=3-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-x=4\\3x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\).
1, \(\left(1,5.x-\frac{4}{5}\right).\left(\frac{1}{2019}-\frac{1}{2018}\right)\)\(=0\)
\(\Leftrightarrow\) \(1,5.x-\frac{4}{5}=0:\left(\frac{1}{2019}-\frac{1}{2018}\right)\)
\(1,5.x-\frac{4}{5}=0\)
\(1,5.x=0+\frac{4}{5}\)
\(1,5.x=\frac{4}{5}\)
\(x=\frac{4}{5}:1,5\)
\(x=\frac{4}{5}:\frac{15}{10}\)
\(x=\frac{4}{5}.\frac{10}{15}\)
\(\Rightarrow x=\frac{8}{15}\)
2, \(\frac{2x}{3}+\frac{1}{3}=\left|-\frac{2}{5}\right|\)
\(\Leftrightarrow\frac{2x+1}{3}=\frac{2}{5}\)
\(2x+1=\frac{2}{5}.3\)
\(2x+1=\frac{6}{5}\)
\(2x=\frac{6}{5}-1\)
\(2x=\frac{1}{5}\)
\(x=\frac{1}{5}:2\)
\(x=\frac{1}{5}.\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{10}\)
