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a) vì \(\left|x+\frac{15}{19}\right|\ge0\text{ }\forall\text{ }x\)
\(\Rightarrow\)Mmin \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = \(\frac{-15}{19}\)
b) vì \(\left|x-\frac{4}{7}\right|\ge0\text{ }\forall\text{ }x\)
\(\Rightarrow\)\(\left|x-\frac{4}{7}\right|-\frac{1}{2}\ge\frac{-1}{2}\)
\(\Rightarrow\)Nmin \(\Leftrightarrow\)N = \(\frac{-1}{2}\)\(\Rightarrow\)\(x=\frac{4}{7}\)
a) vì | x + 15/19 | \(\ge\)0 \(\forall\)x
\(\Rightarrow\)Mmin \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = -15/19
b) vì | x - 4/7 | \(\ge\)0 \(\forall\)x
\(\Rightarrow\)|x - 4/7 | - 1/2 \(\ge\)-1/2
\(\Rightarrow\)Nmin \(\Leftrightarrow\)N = -1/2 \(\Rightarrow\)x = 4/7
a, Ta có: \(\left(x-1\right)^4\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
\(\Rightarrow M=\left(x-1\right)^4+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
Vậy \(M_{min}=\dfrac{1}{4}\Leftrightarrow x=1\)
b, Ta có: \(\left(2x-1\right)^2\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
\(\left|y-1\right|\ge0\forall y\)
Dấu "=" xảy ra \(\Leftrightarrow y=1\)
\(\Rightarrow N=3+\left(2x-1\right)^2+\left|y-1\right|\ge3\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
Vậy \(N_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)
\(M=\left|x-\frac{1}{2}\right|+\left|x-1\right|+\left|x+\frac{1}{4}\right|\)
\(+)\left|x-1\right|+\left|x+\frac{1}{4}\right|=\left|1-x\right|+\left|x+\frac{1}{4}\right|\ge\left|1-x+x+\frac{1}{4}\right|=\frac{5}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(1-x\right)\left(x+\frac{1}{4}\right)\ge0\Leftrightarrow-\frac{1}{4}\le x\le1\)
\(+)\left|x-\frac{1}{2}\right|\ge0\).Dấu '=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
\(\Rightarrow M\ge\frac{5}{2}+0=\frac{5}{2}\)
\(\Rightarrow M_{min}=\frac{5}{2}\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le1\\x=\frac{1}{2}\end{cases}\Rightarrow x=\frac{1}{2}}\)
Cảm ơn các bạn nhiều nha