a: \(A=3\left|2x-1\right|-5\ge-5\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
Bài 7:
\(a,A=3\left|2x-1\right|-5\ge-5\\ A_{min}=-5\Leftrightarrow x=\dfrac{1}{2}\\ b,B=\left|x-1\right|+\left(y+2\right)^2+2021\ge2021\\ B_{min}=2021\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Bài 8:
\(a,C=10-5\left|x-2\right|\le10\\ C_{max}=10\Leftrightarrow x=2\\ b,D=\dfrac{45}{\left(3x-1\right)^2+9}\le\dfrac{45}{0+9}=5\\ D_{max}=5\Leftrightarrow x=\dfrac{1}{3}\)
Bài 9:
\(a,\) Áp dụng tc dtsbn:
\(\dfrac{2x-3}{5}=\dfrac{3y+2}{7}=\dfrac{z-1}{3}=\dfrac{4x-6-6y-4+5z-5}{5\cdot2-7\cdot2+3\cdot5}=\dfrac{33}{11}=3\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=15\\3y+2=21\\z-1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=\dfrac{19}{3}\\z=10\end{matrix}\right.\)
a) \(A=3\left|2x-1\right|-5\)
Vì \(3\left|2x-1\right|\text{≥}0\)
⇒A≥5
Min A=5 ⇔ \(x=\dfrac{1}{2}\)
b) \(B=\left|x-1\right|+\left(y+2\right)^2+2021\)
Vì \(\left\{{}\begin{matrix}\left|x-1\right|\text{≥}0\\\left(y+2\right)^2\text{≥}0\end{matrix}\right.\)
⇒ B ≥2021
Min B=2021⇔ \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
