Tìm max của B=\(\dfrac{x^2}{x^4+x^2+1}\)
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
24 tháng 6 2021
a) đk: x\(\ge0\);
P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P = \(\dfrac{8}{9}\)
<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)
<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)
<=> \(-2x+5\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
c)
Đặt \(\sqrt{x}=a\) (\(a\ge0\))
P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)
Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)
Dấu "=" <=> a = -1 (loại)
=> Không tìm được Min của P
Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)
<=> \(P\le\dfrac{4}{3}\)
Dấu "=" <=> a = 1 <=> x = 1 (tm)
29 tháng 11 2021
\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân VTV
\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)
29 tháng 11 2021
\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)
6 tháng 9 2021
Ta có \(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Leftrightarrow xy\ge1\)
\(A=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{x^2+y^4+2x^2y}\\ \le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}=\dfrac{1}{4xy}+\dfrac{1}{4xy}\\ \le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow x=y=1\)
HN
Cho \(x,y\ne0\) thỏa mãn \(2x^2+\dfrac{1}{x^2}+\dfrac{y^4}{4}=4\) .
Tìm MIN, MAX của : P= \(xy+2021\)
2
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2023
Em kiểm tra đề là \(\dfrac{y^2}{4}\) hay \(\dfrac{y^4}{4}\)
Nếu đề đúng là \(\dfrac{y^4}{4}\) thì có thể coi như là không giải được
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2023
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2-xy+\dfrac{y^2}{4}\right)+xy=2\)
\(\Leftrightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2+xy\ge xy\)
\(\Rightarrow P_{max}=2023\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;-2\right);\left(1;2\right)\)
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2+xy+\dfrac{y^2}{4}\right)-xy=2\)
\(\Rightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x+\dfrac{y}{2}\right)^2-xy\ge-xy\)
\(\Rightarrow xy\ge-2\Rightarrow P\ge2019\)
\(P_{min}=2019\) khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x+\dfrac{y}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;2\right);\left(1;-2\right)\)
2 tháng 12 2021
Câu 1:
a, Giả sử \(A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{a}\ge0\)
\(\Leftrightarrow A=\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge0\)
Mà \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow A\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2\cdot\dfrac{a}{b}-2\cdot\dfrac{b}{a}+2\ge0\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}-2\cdot\dfrac{a}{b}+1\right)+\left(\dfrac{b^2}{a^2}-2\cdot\dfrac{b}{a}+1\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{a}-1\right)^2\ge0\left(\text{luôn đúng}\right)\)
Dấu \("="\Leftrightarrow a=b\)
b, \(B=\dfrac{a^4}{b^4}+\dfrac{b^4}{a^4}-2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\right)+2+\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-4\)
\(B=\left(\dfrac{a^4}{b^4}-2\cdot\dfrac{a^2}{b^2}+1\right)+\left(\dfrac{b^4}{a^4}-2\cdot\dfrac{b^2}{a^2}+1\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)-2\\ \Leftrightarrow B=\left(\dfrac{a^2}{b^2}-1\right)^2+\left(\dfrac{b^2}{a^2}-1\right)^2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2+\dfrac{a}{b}+\dfrac{b}{a}-4\\ \Leftrightarrow B\ge0+0+0+\dfrac{a^2+b^2}{ab}-4\ge\dfrac{2ab}{ab}-4=2-4=-2\)
Dấu \("="\Leftrightarrow\left(a;b\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)
Câu 2:
\(\left(x^2+y^2\right)\left(3^2+4^2\right)\ge\left(3x+4y\right)^2=M^2\\ \Leftrightarrow M^2\le25\cdot25\\ \Leftrightarrow M\le25\)
Dấu \("="\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{25}{25}=1\Leftrightarrow\left\{{}\begin{matrix}x^2=9\\y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy \(M_{max}=25\Leftrightarrow\left(x;y\right)=\left(3;4\right)\)
23 tháng 1 2021
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
XO
27 tháng 12 2021
a) ĐKXĐ : \(3\le x\le7\)
Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)
\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)
Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)
\(B=\dfrac{3x^2}{3\left(x^4+x^2+1\right)}=\dfrac{x^4+x^2+1-x^4+2x^2-1}{3\left(x^4+x^2+1\right)}=1-\dfrac{\left(x^2-1\right)^2}{3\left(x^4+x^2+1\right)}\le1\)
Dấu "=" xảy ra khi \(x=\pm1\)