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a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)
Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).
\(---\)
b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).
\(---\)
c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)
\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)
Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).
a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra khi:
\(\dfrac{2}{5}-x=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
Vậy: ...
b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)
Dấu "=" xảy ra:
\(x+\dfrac{2}{3}=0\)
\(\Rightarrow x=-\dfrac{2}{3}\)
Vậy: ...
c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)
Dấu "=" xảy ra:
\(\dfrac{7}{4}-x=0\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy: ...
3 câu này bạn áp dụng cái này nhé.
`a^2 >=0 forall a`.
`|a| >=0 forall a`.
`1/a` xác định `<=> a ne 0`.
a: P=(x+30)^2+(y-4)^2+1975>=1975 với mọi x,y
Dấu = xảy ra khi x=-30 và y=4
b: Q=(3x+1)^2+|2y-1/3|+căn 5>=căn 5 với mọi x,y
Dấu = xảy ra khi x=-1/3 và y=1/6
c: -x^2-x+1=-(x^2+x-1)
=-(x^2+x+1/4-5/4)
=-(x+1/2)^2+5/4<=5/4
=>R>=3:5/4=12/5
Dấu = xảy ra khi x=-1/2
a) \(A=\left(x-1\right)^2\ge0\)
Dấu " = " xảy ra :
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(Min_A=0\Leftrightarrow x=1\)
b) Ta thấy : \(\left(x^2-9\right)^2\ge0\)
\(\left|y-2\right|\ge0\)
\(\Leftrightarrow B=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)
Vậy \(Min_B=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)
c) Ta thấy : \(x^4\ge0\)
\(x^2\ge0\)
\(\Leftrightarrow C=x^4+3x^2+2\ge2\)
Dấu " = " xảy ra ;
\(\Leftrightarrow x=0\)
Vậy \(Min_C=2\Leftrightarrow x=0\)
d) \(D=x^2+4x-100\)
\(\Leftrightarrow D=x^2+4x+4-104\)
\(\Leftrightarrow D=\left(x+2\right)^2-104\ge-104\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(Min_D=-104\Leftrightarrow x=-2\)
Bài 1 :
a) Ta thấy : \(\left(x^2-9\right)^2\ge0\)
\(\left|y-2\right|\ge0\)
\(\Leftrightarrow A=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)
Vậy \(Min_A=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)
b) Ta thấy : \(B=x^2+4x-100\)
\(=\left(x+4\right)^2-104\ge-104\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy \(Min_B=-104\Leftrightarrow x=-4\)
c) Ta thấy : \(C=\frac{4-x}{x-3}\)
\(=\frac{3-x+1}{x-3}\)
\(=-1+\frac{1}{x-3}\)
Để C min \(\Leftrightarrow\frac{1}{x-3}\)min
\(\Leftrightarrow x-3\)max
\(\Leftrightarrow x\)max
Vậy để C min \(\Leftrightarrow\)\(x\)max
p/s : riêng câu c mình không tìm được C min :( Mong bạn nào giỏi tìm hộ mình
Bài 2 :
a) Ta thấy : \(x^2\ge0\)
\(\left|y+1\right|\ge0\)
\(\Leftrightarrow3x^2+5\left|y+1\right|-5\ge-5\)
\(\Leftrightarrow C=-3x^2-5\left|y+1\right|+5\le-5\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy \(Max_A=-5\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)
b) Để B max
\(\Leftrightarrow\left(x+3\right)^2+2\)min
Ta thấy : \(\left(x+3\right)^2\ge0\)
\(\Leftrightarrow\left(x+3\right)^2+2\ge2\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(Max_B=\frac{1}{2}\Leftrightarrow x=-3\)
c) Ta thấy : \(\left(x+1\right)^2\ge0\)
\(\Leftrightarrow x^2+2x+1\ge0\)
\(\Leftrightarrow-x^2-2x-1\le0\)
\(\Leftrightarrow C=-x^2-2x+7\le8\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(Max_C=8\Leftrightarrow x=-1\)
2:
a: =-(x^2-12x-20)
=-(x^2-12x+36-56)
=-(x-6)^2+56<=56
Dấu = xảy ra khi x=6
b: =-(x^2+6x-7)
=-(x^2+6x+9-16)
=-(x+3)^2+16<=16
Dấu = xảy ra khi x=-3
c: =-(x^2-x-1)
=-(x^2-x+1/4-5/4)
=-(x-1/2)^2+5/4<=5/4
Dấu = xảy ra khi x=1/2
1)
a) \(A=x^2+4x+17\)
\(A=x^2+4x+4+13\)
\(A=\left(x+2\right)^2+13\)
Mà: \(\left(x+2\right)^2\ge0\) nên \(A=\left(x+2\right)^2+13\ge13\)
Dấu "=" xảy ra: \(\left(x+2\right)^2+13=13\Leftrightarrow x=-2\)
Vậy: \(A_{min}=13\) khi \(x=-2\)
b) \(B=x^2-8x+100\)
\(B=x^2-8x+16+84\)
\(B=\left(x-4\right)^2+84\)
Mà: \(\left(x-4\right)^2\ge0\) nên: \(A=\left(x-4\right)^2+84\ge84\)
Dấu "=" xảy ra: \(\left(x-4\right)^2+84=84\Leftrightarrow x=4\)
Vậy: \(B_{min}=84\) khi \(x=4\)
c) \(C=x^2+x+5\)
\(C=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)
\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)
Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu "=" xảy ra: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{1}{2}\)
a) |x+2|+|3-x|>=|x+2+3-x|=|5|=5
dau "=" xay ra khi va chi khi (x+2)(3-x)>=0
=>x>=-2 hoặc x<=3
vạy GTNN cua bieu thuc la 5 khi va chi khi ...
b)cau b tuong tu
c) vi |x+1|>=0
|y+2|>=0
=>|x+1|+|y+2|>=0 dau "=" xay ra khi va chi khi x+1=0 va y+2=0
=>x=-1 va y=-2
vay GTNN cua bieu thuc la 0 khi va chi khi x=-1 va y=-2