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\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\)
b.\(B=7-\left(x+3\right)^2\le7\forall x\) " = " \(\Leftrightarrow x=-3\)
c.\(C=\left|2x-3\right|-13\ge-13\forall x\) " = " \(\Leftrightarrow x=\dfrac{3}{2}\)
d.\(D=11-\left|2x-13\right|\le11\forall x\) " = " \(\Leftrightarrow x=\dfrac{13}{2}\)
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\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)
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Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
$A=|x-2001|+|x-1|=|2001-x|+|x-1|\geq |2001-x+x-1|=2000$
Vậy $A_{\min}=2000$. Giá trị này đạt được khi $(2001-x)(x-1)\geq 0$
$\Leftrightarrow 2001\geq x\geq 1$
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Vì |1/2 - x| > 0
=> 0,6 + |1/2 - x| > 0,6
=> A > 0,6
Dấu "=" xảy ra
<=> 1/2 - x = 0
<=> x = 1/2
KL: Amin = 0,6 <=> x = 1/2
Vì |2x + 2/3| > 0
=> 2/3 - |2x + 2/3| < 2/3
=> B < 2/3
Dấu "=" xảy ra
<=> 2x + 2/3 = 0
<=> 2x = -2/3
<=> x = -1/3
KL: Bmax = 2/3 <=> x = -1/3
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Ta có:\(M=\left|x-2002\right|+\left|x-2001\right|\)
\(=\left|2002-x\right|+\left|x-2001\right|\ge\left|2002-x+x-2001\right|=\left|1\right|=1\)
Vậy \(MinM=1\) khi \(\orbr{\begin{cases}x=2002\\x=2001\end{cases}}\)
Áp dụng đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|.\) dấu = khi \(AB\ge0\)
Mà \(M=\left|x-2002\right|+\left|x-2001\right|=\left|x-2002\right|+\left|2001-x\right|\)
\(\Rightarrow M=\left|x-2002\right|+\left|2001-x\right|\ge\left|x-2002+2001-x\right|\)
\(\Rightarrow M\ge\left|-1\right|\Rightarrow M\ge1\)dấu = khi \(\left(x-2002\right)\left(2001-x\right)\ge0\)
Vậy \(M_{min}=1\)
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a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)
\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)
b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)
\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
A,B,C riêng nha
A=x2−4x+1=(x−2)2−3≥−3A=x2−4x+1=(x−2)2−3≥−3
⇒Amin=−3⇒Amin=−3 khi x=2x=2
B=4x2+4x+11=(2x+1)2+10≥10B=4x2+4x+11=(2x+1)2+10≥10
⇒Bmin=10⇒Bmin=10 khi x=−12x=−12
C=(x−1)(x+6)(x+2)(x+3)=(x2+5x−6)(x2+5x+6)C=(x−1)(x+6)(x+2)(x+3)=(x2+5x−6)(x2+5x+6)
=(x2+5x)2−36≥−36=(x2+5x)2−36≥−36
⇒Cmin=−36⇒Cmin=−36 khi [x=0x=−5[x=0x=−5
D=−x2−8x−16+21=21−(x+4)2≤21D=−x2−8x−16+21=21−(x+4)2≤21
⇒Cmax=21⇒Cmax=21 khi x=−4x=−4
E=−x2+4x−4+5=5−(x−2)2≤5E=−x2+4x−4+5=5−(x−2)2≤5
⇒Emax=5⇒Emax=5 khi x=2