a,Tim GTNN cua bieu thuc \(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
b,Tim GTLN cua bieu thuc \(D=\frac{4}{\left(2x-3\right)^2+5}\)
Những câu hỏi liên quan
Tim GTNN hoac GTLN cua cac bieu thuc sau :
a) A = 3|2x - 1| - 5 b) B = 10 - 5 |x - 2| c) C =\(\frac{1}{\left|x-2\right|+3}\)
Tim GTNN cua bieu thuc:
C=\(\frac{-2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)
Ta có :
\(C=-\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)
Ta có : | x + 4 | \(\ge\)0 ; ( y - 1.3 )104 \(\ge\)0
\(\Rightarrow\) | x + 4 | + ( y - 1.3 )104 \(\ge\)0
\(\Rightarrow\)| x + 4 | + ( y - 1.3 )104 + 18 \(\ge\)18
Dấu " = " xảy ra khi \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)
\(\Rightarrow\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\le\frac{2}{18}=\frac{1}{9}\)
\(\Rightarrow\)GTLN của \(\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)là \(\frac{1}{9}\)
\(\Rightarrow\)\(-\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)có GTNN của \(\frac{1}{9}\)
Vậy Cmin = \(\frac{1}{9}\)khi \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)
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cau 1: tinh gia tri cua x thoa manleft(x-3right)left(x^2+3x+9right)+xleft(x+2sqrt{2}right)left(2sqrt{2}-xright)-3cau 2.tinh GTLN cua bieu thuc2x-2x^2+13cau 3. tinh gia tri cua bieu thuc frac{3^{left(x+yright)^2}}{3^{left(x-yright)^2}}voi xyfrac{1}{2}cau 4. tim GTLN cua-3x^2-6x-4cau 5. cho ham so : f(x)frac{1}{5x+9}tinh gia tri cua fleft(frac{40}{25}right)cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
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cau 1: tinh gia tri cua x thoa man
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\sqrt{2}\right)\left(2\sqrt{2}-x\right)=-3\)
cau 2.tinh GTLN cua bieu thuc
\(2x-2x^2+13\)
cau 3. tinh gia tri cua bieu thuc
\(\frac{3^{\left(x+y\right)^2}}{3^{\left(x-y\right)^2}}\)voi xy=\(\frac{1}{2}\)
cau 4. tim GTLN cua
\(-3x^2-6x-4\)
cau 5. cho ham so : f(x)=\(\frac{1}{5x+9}\)
tinh gia tri cua \(f\left(\frac{40}{25}\right)\)
cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
Cho bieu thuc: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
tim GTLN cua bieu thuc :\(A=\left(1-x^n\right)\left(1+x^n\right)+\left(2-y^n\right)\left(2+y^n\right)\)
\(A=\left(1-x^{2n}\right)+\left(2-y^{2n}\right)\)
Có \(x^{2n}\ge0\);\(y^{2n}\ge0\)
\(\Rightarrow A\le\left(1-0\right)+\left(2-0\right)=3\)
Dấu "=" xảy ra khi x = 0 ; y = 0 với mọi n
Vậy Max A = 3 <=> x = 0 ; y = 0
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Tim gia tri cua bieu thuc A=\(3\left(\frac{2}{\sqrt{10}+5}+\frac{5}{\sqrt{10}-2}-\frac{7}{\sqrt{10}}\right)\)
Tim GTLN cua bieu thuc: Q = \(\frac{-\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\)
ta có : (\(\sqrt{x}\)- 2 )\(^2\)\(\ge\)0
\(\Leftrightarrow\)x - 4\(\sqrt{x}\)+ 4 \(\ge\)0
\(\Leftrightarrow\)x - 4\(\sqrt{x}\)+ 4 + 8\(\sqrt{x}\) \(\ge\)8\(\sqrt{x}\)
\(\Leftrightarrow\)(\(\sqrt{x}\)+ 2 )\(^2\)\(\ge\)8\(\sqrt{x}\)
\(\Leftrightarrow\)-(\(\sqrt{x}\)+ 2 )\(^2\)\(\le\)-8\(\sqrt{x}\)
\(\Leftrightarrow\)Q \(\le\)\(\frac{-8\sqrt{x}}{\sqrt{x}}\)= ( - 8 )
Dấu '' = '' xaye ra tại x = 4
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tim GTLN,GTNN cua bieu thuc sau
D=\(\left|4x-3\right|+\left|5y+7,5\right|+17,5\)
E=\(4-\left|5x-2\right|-\left|3y+12\right|\)
\(\)bài nào có MIN or MAX thì mk làm,mk ko làm thì có nghĩa là ko có nha
\(D=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)
\(\left\{{}\begin{matrix}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|\ge0\)
\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|4x-3\right|=0\Rightarrow4x=3\Rightarrow x=\dfrac{3}{4}\\\left|5y+7,5\right|=0\Rightarrow5y=-7,5\Rightarrow y=-1,5\end{matrix}\right.\)
\(\Rightarrow MIN_D=17,5\) khi \(x=\dfrac{3}{4};y=-1,5\)
\(E=4-\left|5x-2\right|-\left|3y+12\right|\)
\(\left\{{}\begin{matrix}\left|5x-2\right|\ge0\\\left|3y+12\right|\ge0\end{matrix}\right.\)
\(\Rightarrow E=4-\left|5x-2\right|-\left|3y+12\right|\le4\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|5x-2\right|=0\Rightarrow5x=2\Rightarrow x=\dfrac{2}{5}\\\left|3y+12\right|=0\Rightarrow3y=-12\Rightarrow y=-4\end{matrix}\right.\)
\(\Rightarrow MAX_E=4\) khi \(x=\dfrac{2}{5};y=-4\)
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