Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-1\right)^2\ge0;\left|2y+2\right|\ge0\Rightarrow\left(x-1\right)^2+\left|2y+2\right|-3\ge-3\)
dấu = xảy ra khi \(\hept{\begin{cases}x-1=0\\2y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
vậy GTNN của C là -3 khi x=1, y=-1
a)\(9-3x< 0\)
\(9< 3x\)
\(x>3\)
b)\(\left(2x-4\right)\left(x+5\right)< 0\)
\(\Leftrightarrow2x-4>0\Rightarrow2x>4\Rightarrow x>2\)
\(x+5< 0\Rightarrow x< -5\)
\(2< x< -5\)
\(x\in\left\{\varnothing\right\}\)
\(\Leftrightarrow2x-4< 0\Rightarrow2x< 4\Rightarrow x< 2\)
\(x+5>0\Rightarrow x>-5\)
\(-5< x< 2\)
\(x\in\left\{-4;-3;-2;-1;0;1\right\}\)
c)
\(\dfrac{x+4}{x-2}\in Z^-\)
\(\Leftrightarrow x+4< 0\Rightarrow x< -4\)
\(x-2>0\Rightarrow x>2\)
\(2< x< -4\)
\(x\in\left\{\varnothing\right\}\)
\(\Leftrightarrow x+4>0\Rightarrow x>-4\)
\(x-2< 0\Rightarrow x< 2\)
\(-4< x< 2\)
\(x\in\left\{-3;-2;-1;0;1\right\}\)
a) Vì \(\left|2x+8\right|\ge0\forall x\)
\(\Rightarrow\left|2x+8\right|-3\ge-3\forall x\)
\(\Rightarrow A_{min}=-3\)
Dấu "=" xảy ra khi: \(2x+8=0\)
\(\Leftrightarrow2x=-8\)
\(\Leftrightarrow x=-4\left(TM\right)\)
Vậy \(A_{min}=-3\)\(\Leftrightarrow\)\(x=-4\)
a) M=-(x-2)2
ta có (x-2)2 >=0 với mọi x
=> -(x-2)2 =<0. Dấu "=" xảy ra <=> (x-2)2=0
<=> x-2=0
<=> x=2
Vậy MaxM=0 đạt được khi x=2
b) Ta có |x+5| >=0 với mọi x
=> -|x+5| =<0 => -|x+5|-2 =<-2
Dấu "=" xảy ra <=> |x+5|=0
<=> x=-5
Vậy MaxN=-2 đạt được khi x=-5
\(a,\frac{x-3}{x+4}=\frac{x+4-7}{x+4}=1-\frac{7}{x+4}\\ \Rightarrow x+4\inƯ\left(7\right)=\left\{-1;-7;1;7\right\}\)
\(b,\frac{3x-15}{x-4}=\frac{3x-12-3}{x-4}=3-\frac{3}{x-4}\\ \Rightarrow x-4\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(c,\frac{2x+11}{x+3}=\frac{2x+6+5}{x+3}=2+\frac{5}{x+3}\\ \Rightarrow x+3\inƯ\left(5\right)=\left\{-1;5;-5;1\right\}\)
\(d,\frac{x+5}{x-2}=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\\ \Rightarrow x-2\inƯ\left(7\right)=\left\{-1;-7;1;7\right\}\)
a) Để \(\frac{7-x}{x-2}\inℤ\) thì \(\left(7-x\right)⋮\left(x-2\right)\)
\(\Leftrightarrow\left[-1\left(7-x\right)\right]⋮\left(x-2\right)\)
\(\Leftrightarrow\left[x-7\right]⋮\left(x-2\right)\)
\(\Leftrightarrow\left[x-2-5\right]⋮\left(x-2\right)\)
Vì \(\Leftrightarrow\left[x-2\right]⋮\left(x-2\right)\) nên \(\Leftrightarrow5⋮\left(x-2\right)\)
hay \(x-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Lập bảng:
| \(x-2\) | \(1\) | \(-1\) | \(5\) | \(-5\) |
| \(x\) | \(3\) | \(1\) | \(7\) | \(-3\) |
Vậy \(x\in\left\{1;\pm3;7\right\}\)
b) Để \(\frac{x+8}{3-x}\inℤ\) thì \(\left(x+8\right)⋮\left(3-x\right)\)
\(\Leftrightarrow\left[-1\left(x+8\right)\right]⋮\left(3-x\right)\)
\(\Leftrightarrow\left[8-x\right]⋮\left(3-x\right)\)
\(\Leftrightarrow\left[5+3-x\right]⋮\left(3-x\right)\)
Vì \(\left[3-x\right]⋮\left(3-x\right)\) nên \(5⋮\left(3-x\right)\)
Lập bảng như câu a)
a) \(A=\left(x-1\right)^2+2004\)
Vì \(\left(x-1\right)^2\ge0\) nên \(A=\left(x-1\right)^2+2004\ge2004\)
\(\Rightarrow A_{min}\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=0+1\)
\(\Leftrightarrow x=1\)
Vậy Amin = 2014 \(\Leftrightarrow x=1\)
b) \(B=\left|x+4\right|+2014\)
Vì \(\left|x+4\right|\ge0\) nên \(B=\left|x+4\right|+2014\ge2014\)
\(\Rightarrow B_{min}\Leftrightarrow\left|x+4\right|=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=0-4\)
\(\Leftrightarrow x=-4\)
Vậy Bmin = 2014\(\Leftrightarrow x=-4\)