Đăng một bài easy cho tất cả mọi người.
Tìm Max của biểu thức sau: \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DH
5 tháng 1 2020
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x-2}}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)
a, \(Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x-2}}\right).\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}+2}.\frac{1}{\sqrt{x}}\)
\(=\frac{2}{\sqrt{x}+2}\)
\(b,x>0;x\ne4\)
\(A>\frac{1}{2}\)
\(\Rightarrow\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)
\(\Rightarrow4>\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}< 2\)
\(\Rightarrow x< 4\)
Vậy \(0< x< 4\)
c, \(B=\frac{5}{2}.A=\frac{5}{2}.\frac{2}{\sqrt{x}+2}=\frac{5}{\sqrt{x}+2}\)
\(B\in Z\)
\(\Rightarrow\left(\sqrt{x}+2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Vì: \(\sqrt{x}+2>2\forall x>0;x\ne4\)
\(\Rightarrow\sqrt{x}+2=5\)
\(\Rightarrow\sqrt{x}=3\)
\(\Rightarrow x=9\left(tm\right)\)
Vậy ...................................................
14 tháng 5 2020
\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)
a) \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)
\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)
\(A=\frac{2x^2+x-1+x^2+4x.4}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)^2}\)
Ta có :\(3x^2+5x+3\)
\(=3\left(x^2+\frac{5}{3}x+1\right)\)
\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)
\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)
Mà \(\left(x-2\right)^2>0\)
\(\Rightarrow A>0\left(dpcm\right)\)
\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)
\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)
\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)
\(\Rightarrow8x^2-49x+41=0\)
\(\Rightarrow8x^2-8x-41x+41=0\)
\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)
\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)
MN
4 tháng 4 2020
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
TM
10 tháng 7 2017
a)cần điều kiện xác định thì bạn tự tìm
\(A=\left(\frac{1}{x+2}+\frac{1}{x-2}\right).\frac{x-2}{x}=\left(\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right).\frac{x-2}{x}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x}=\frac{2}{x+2}\)
b)\(A=\frac{2}{x+2}>\frac{1}{2}\Leftrightarrow4>x+2\Leftrightarrow x< 2\)
c)\(B=\frac{7}{3}A=\frac{7}{3}.\frac{2}{x+2}=\frac{14}{3x+6}\)
B nguyên khi 14 chia hết cho 3x+6 <=> 3x+6 \(\inƯ\left(14\right)=\left\{-14;-7;-2;-1;1;2;7;14\right\}\)
<=>\(3x\in\left\{-20;-13;-8;-7;-5;-4;1;8\right\}\)
<=>\(3x\in\left\{1;8\right\}\) do x dương => 3x dương
<=>x\(\in\left\{\frac{1}{3};\frac{8}{3}\right\}\)
5 tháng 7 2017
Bài 2 :
Ta có : \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\ge\frac{3}{4}\forall x\in R\)
Vậy Amin = \(\frac{3}{4}\) dấu "=" chỉ sảy ra khi x = \(\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 4 2022
\(\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)\ge\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)^2\)
\(\Rightarrow\sqrt{a^2+\dfrac{1}{b+c}}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)=\dfrac{1}{\sqrt{17}}\left(4a+\dfrac{1}{\sqrt{b+c}}\right)\)
Tương tự:
\(\sqrt{b^2+\dfrac{1}{a+c}}\ge\dfrac{1}{\sqrt{17}}\left(4b+\dfrac{1}{\sqrt{a+c}}\right)\) ; \(\sqrt{c^2+\dfrac{1}{a+b}}\ge\dfrac{1}{\sqrt{17}}\left(4c+\dfrac{1}{\sqrt{a+b}}\right)\)
Cộng vế:
\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)
Cũng theo Bunhiacopxki:
\(1.\sqrt{a+b}+1.\sqrt{b+c}+1\sqrt{c+a}\le\sqrt{\left(1+1+1\right)\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)
\(\Rightarrow VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\)
\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=2\)
cho ít bài lớp 8 cm hình đi
Áp dụng bất đẳng thức Côsy cho các cặp số không âm (a^2,1);(b^2,1),(c^2,1) ta có: a^2 +1 >= 2a ; b^2 + 1 >= 2b ; c^2 + 1 >= 2c
Do đó: \(\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\le\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}=\frac{3}{2}\)
Đẳng thức xảy ra <=> a^2 = 1 ; b^2 = 1 ; c^2 = 1 <=> \(\hept{\begin{cases}a=\pm1\\b=\pm1\\c=\pm1\end{cases}}\)