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a = |2x-1/3|-7/4
Do |2x-1/3| \(\ge\) 0
|2x-1/3|-7/4 \(\ge\) 7/4
Dấu = xảy ra <=> 2x-1/3=0. =>. x= 1/6
b 1/3|x-2|+2|3-1/2 y|+4
Do |x-2| \(\ge\) 0
|3-1/2y| \(\ge\) 0
=> 1/3|x-2|+2|3-1/2 y|+4 \(\ge\) 4
Dấu = xảy ra <=>\(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
a: Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{6}\)
b: Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(2\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+2\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\left|x-2\right|\cdot\dfrac{1}{3}+\left|3-\dfrac{1}{2}y\right|\cdot2+4\ge4\forall x,y\)
Dấu '=' xảy ra khi x=2 và y=6
\(-x-2=\dfrac{-5}{4}\\ \Leftrightarrow x=-2+\dfrac{5}{4}=-\dfrac{3}{4}\)
\(\left(x+\dfrac{1}{2}\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=2\\x+\dfrac{1}{2}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(\dfrac{x}{2}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{20}=\dfrac{z}{12}\)
Áp dụng t/c của dãy số bằng nhau, ta có: \(\dfrac{x-y+z}{10-20+12}=\dfrac{4}{2}=2\)
\(\dfrac{x}{10}=2\Rightarrow x=20\)
\(\dfrac{y}{20}=2\Rightarrow y=40\)
\(\dfrac{z}{12}=2\Rightarrow z=24\)
x/10=y/20=z/12
x-y+z/=10-20+12=4/2=2
x=2.10=20
y=2.20=40
z=2.12=24
\(\dfrac{3}{4}:\left(2\dfrac{4}{9}\right)-\left|-3x+2\dfrac{2}{3}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{9}{22}\)
\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{27}{88}=\dfrac{39}{88}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{8}{3}=\dfrac{39}{88}\\3x-\dfrac{8}{3}=-\dfrac{39}{88}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{821}{792}\\x=\dfrac{587}{792}\end{matrix}\right.\)
\(\dfrac{3^8\cdot20^5-3^9\cdot5^5\cdot2^9}{6^8\cdot10^4-3^8\cdot2^9\cdot5^4}=\dfrac{3^8\cdot2^{10}\cdot5^5-3^9\cdot5^5\cdot2^9}{2^8\cdot3^8\cdot2^4\cdot5^4-3^8\cdot2^9\cdot5^4}\\ =\dfrac{3^8\cdot2^9\cdot5^5\left(2-3\right)}{2^9\cdot3^8\cdot5^4\left(2^3-1\right)}=\dfrac{-5}{2^3-1}=\dfrac{-5}{7}\)
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(x+\frac{4}{12}-1+\frac{2}{3}=-2+\frac{3}{4}\)
\(\Leftrightarrow\frac{12x+4-12+8}{12}=\frac{-24+9}{12}\)
\(\Leftrightarrow12x=-15\)
\(\Leftrightarrow x=-\frac{5}{4}\)
Tìm x
\(x+\frac{4}{12}-1+\frac{2}{3}=-2+\frac{3}{4}\)
\(x+\frac{1}{3}-1+\frac{2}{3}=-\frac{8}{4}+\frac{3}{4}\)
\(x+\left(\frac{1}{3}-1+\frac{2}{3}\right)=-\frac{5}{4}\)
\(x+\left(\frac{1}{3}-\frac{3}{3}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(x+0=-\frac{5}{4}\)
\(x=-\frac{5}{4}\)
Vậy \(x=-\frac{5}{4}\).