Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{a^4}{a\sqrt{b^2+3}}+\frac{b^4}{b\sqrt{a^2+3}}+\frac{c^4}{c\sqrt{a^2+3}}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{b^2+3}+b\sqrt{a^2+3}+c\sqrt{a^2+3}}\)
Ta co : \(2a\sqrt{b^2+3}+2b\sqrt{c^2+3}+2c\sqrt{a^2+3}\le\frac{4a^2+b^2+3}{2}+\frac{4b^2+a^2+3}{2}+\frac{4a^2+c^2+3}{2}\)
=> \(2\left(a\sqrt{b^2+3}+b\sqrt{c^2+3}+c\sqrt{a^2+3}\right)\le\frac{5\left(a^2+b^2+c^2\right)+9}{2}\) = \(\frac{5.3+9}{2}=12\)
=> \(a\sqrt{b^2+3}+b\sqrt{c^2+3}+c\sqrt{a^2+3}\le6\)
=> \(P\ge\frac{\left(a^2+b^2+c^2\right)^2}{6}=\frac{9}{6}=\frac{3}{2}\)
Dấu '' = '' xảy ra khi và chỉ khi : \(\frac{a^2}{\sqrt{b^2+3}}=\frac{b^2}{\sqrt{c^2+3}}=\frac{c^2}{\sqrt{a^2+3}}\)
\(2a=\sqrt{b^2+3};2b=\sqrt{c^2+3};2c=\sqrt{a^2+3}\)
\(a^2+b^2+c^2=3\)
=> a = b= c = 1
ta có:\(a^2+ab+b^2=\frac{3}{4}\left(a+b\right)^2+\frac{1}{4}\left(a-b\right)^2\ge\frac{3}{4}\left(a+b\right)^2\)
hay \(\sqrt{a^2+ab+b^2}\ge\frac{\sqrt{3}}{2}\left(a+b\right)\)
tương tự ta có:\(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}}{2}\left(b+c\right),\sqrt{c^2+ac+c^2}\ge\frac{\sqrt{3}}{2}\left(a+c\right)\)
đến đây tự full đi
Ta có 4S=\(a\sqrt[3]{8.8.\left(b^2+c^2\right)}+b\sqrt[3]{8.8.\left(c^2+a^2\right)}+c.\sqrt[3]{8.8\left(a^2+b^2\right)}\)
Áp dụng BĐT cô-si, ta có \(\sqrt[3]{8.8.\left(b^2+c^2\right)}\le\frac{8+8+b^2+c^2}{3}\Rightarrow a\sqrt[3]{8.8.\left(b^2+c^2\right)}\le\frac{16}{3}a+\frac{1}{3}\left(ab^2+ac^2\right)\)
Tương tự, rồi cộng lại, ta có
\(4S\le\frac{16}{3}\left(a+b+c\right)+\frac{1}{3}\left(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2\right)\)
Mà \(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=6\)
\(ab^2+bc^2+ca^2+a^2b+b^2c+c^2a\le\frac{2}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{2}{3}.6.12=48\)
=> \(4S\le\frac{16}{3}.6+\frac{1}{3}.48=48\Rightarrow S\le12\)
Dấu = xảy ra <=> a=b=c=2
^_^
Ta có: \(x^2+xy+y^2=\frac{1}{2}\left(x+y\right)^2+\frac{1}{2}\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+y\right)^2+\frac{1}{4}\left(x+y\right)^2=\frac{3}{4}\left(x+y\right)^2\)
\(\Rightarrow M\ge\sqrt{\frac{3}{4}\left(a+b\right)^2}+\sqrt{\frac{3}{4}\left(b+c\right)^2}+\sqrt{\frac{3}{4}\left(c+a\right)^2}\)
\(\Rightarrow M\ge\sqrt{3}\left(a+b+c\right)=3\sqrt{3}\)
\(M_{min}=3\sqrt{3}\) khi \(a=b=c=1\)
Bài 1 :
+) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a) Ta có :
\(x=4-2\sqrt{3}\)
\(\Leftrightarrow x=3-2\sqrt{3}+1\)
\(\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)( Thỏa mãn ĐKXĐ )
Vậy tại \(x=\left(\sqrt{3}-1\right)^2\)thì giá trị của biểu thức A là :
\(A=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}+1}{\sqrt{\left(\sqrt{3}-1\right)^2}-3}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-3}=\frac{\sqrt{3}}{\sqrt{3}-4}=\frac{-\sqrt{3}\left(\sqrt{3}+4\right)}{7}\)
b)
\(B=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{-3-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Ta có :
\(P=A:B\)
\(\Leftrightarrow P=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{-3\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\frac{-\sqrt{x}-3}{3}\)
c) \(P=\frac{-\sqrt{x}-3}{3}\ge0\)
Dấu bằng xảy ra
\(\Leftrightarrow-\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x}=-3\)( vô lí )
Vậy không tìm được giá trị nào của x để P đạt GTNN
a. Từ giả thiết ta có:
\(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow x^2+y^2=4-2xy\ge4-2.\frac{\left(x+y\right)^2}{4}=4-2.\frac{4}{4}=2\)
\(\Rightarrow Min=2\Leftrightarrow x=y=1\)
b. Từ giả thiết suy ra:
\(3\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca\le1\)
\(\Rightarrow T=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\)
\(\le\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{a}{\sqrt{\left(c+b\right)\left(a+b\right)}}+\frac{a}{\sqrt{\left(c+b\right)\left(a+c\right)}}\)
\(=\sqrt{\frac{a}{a+b}.\frac{a}{a+c}}+\sqrt{\frac{b}{c+b}.\frac{b}{a+b}}+\sqrt{\frac{a}{b+c}.\frac{a}{a+c}}\)
\(\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{c+b}+\frac{b}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}\right)\)
\(=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{1}{2}\left(1+1+1\right)=\frac{3}{2}\)
\(Max_T=\frac{3}{2}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}\)
Chắc là tìm MaxP
Có 9=3(a2+b2+c2)\(\ge\)(a+b+c)2
=>a+b+c\(\le\)3
Có P2=\(\left(\sqrt{3+a}+\sqrt{3+b}+\sqrt{3+c}\right)\le\left(1^2+1^2+1^2\right)\left(9+a+b+c\right)\)
=>P2\(\le\)3.12=36
=>P2\(\le\)6
Dấu = xảy ra <=> a2=b2=c2 và \(\frac{\sqrt{3+a}}{1}=\frac{\sqrt{3+b}}{1}=\frac{\sqrt{3+c}}{1}\) và a2+b2+c2=3
<=> a=b=c=1