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Ta có: \(\hept{\begin{cases}|x+1|\ge0;\forall x,y\\2|6,9-3y|\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow|x+1|+2|6,9-3y|\ge0;\forall x,y\)
\(\Rightarrow|x+1|+2|6,9-3y|+3\ge0+3;\forall x,y\)
Hay \(B\ge3;\forall x,y\)
Dấu "=" xảy ra \(\hept{\begin{cases}|x+1|=0\\2|6,9-3y|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2,3\end{cases}}}\)
Vậy MIN \(B=3\Leftrightarrow\hept{\begin{cases}x=-1\\y=2,3\end{cases}}\)
Ta thấy : \(\left|x+1\right|\ge0\forall x\)
\(2\left|6,9-3y\right|\ge0\forall y\)
\(\Rightarrow\left|x+1\right|+2\left|6,9-3y\right|\ge0\forall x,y\)
\(\Rightarrow\left|x+1\right|+2\left|6,9-3y\right|+3\ge3\)
hay \(B\ge3.\) Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=0\\2\left|6.9-3y\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\6,9-3y=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\\6,9=3y\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\\y=2,3\end{cases}}\)
Vậy : B đạt giá trị nhỏ nhất bằng 3 khi \(x=-1;y=2,3\).
2.
a/\(A=5-I2x-1I\)
Ta thấy: \(I2x-1I\ge0,\forall x\)
nên\(5-I2x-1I\le5\)
\(A=5\)
\(\Leftrightarrow5-I2x-1I=5\)
\(\Leftrightarrow I2x-1I=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A=5\Leftrightarrow x=\frac{1}{2}\)
b/\(B=\frac{1}{Ix-2I+3}\)
Ta thấy : \(Ix-2I\ge0,\forall x\)
nên \(Ix-2I+3\ge3,\forall x\)
\(\Rightarrow B=\frac{1}{Ix-2I+3}\le\frac{1}{3}\)
\(B=\frac{1}{3}\)
\(\Leftrightarrow B=\frac{1}{Ix-2I+3}=\frac{1}{3}\)
\(\Leftrightarrow Ix-2I+3=3\)
\(\Leftrightarrow Ix-2I=0\)
\(\Leftrightarrow x=2\)
Vậy GTLN của\(A=\frac{1}{3}\Leftrightarrow x=2\)
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)
a)Ta thấy:
\(-\left|\frac{1}{3}x+2\right|\le0\)
\(\Rightarrow5-\left|\frac{1}{3}x+2\right|\le5-0=5\)
\(\Rightarrow B\le5\)
Dấu "=" xảy ra khi x=-6
Vậy MaxB=5<=>x=-6
b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\).Ta có:
\(\left|\frac{1}{2}x-3\right|+\left|\frac{1}{2}x+5\right|\ge\left|\frac{1}{2}x-3+5-\frac{1}{2}x\right|=2\)
\(\Rightarrow C\ge2\)
Dấu "=" xảy ra khi \(\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)
Vậy MinC=2<=>x=6 hoặc -10
a) Vì (x+2)2 >/ 0
=> \(A\le\frac{3}{0+4}=\frac{3}{4}\Rightarrow Amax=\frac{3}{4}\Leftrightarrow x+2=0\Rightarrow x=-2\)
b) Vì \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\)
\(B\ge0+0+1=1\Rightarrow Bmin=1\Leftrightarrow\int^{x+1=0}_{y+3=0}\Rightarrow\int^{x=-1}_{y=-3}\)
B = |x + 1| + 2|6,9 - 3y| + 3
Nhận thấy \(\hept{\begin{cases}\left|x+1\right|\ge0\forall x\\2\left|6,9-3y\right|\ge0\forall y\end{cases}}\Rightarrow\left|x+1\right|+2\left|6,9-3y\right|\ge0\forall x;y\)
=> \(\left|x+1\right|+2\left|6,9-3y\right|+3\ge3\forall x,y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+1=0\\6,9-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=2,3\end{cases}}\)
Vậy Min B = 3 <=> x = - 1 ; y = 2,3
câu trả lời là 2.3