Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(\sqrt{3b\left(a+2b\right)}\le\frac{1}{2}\left(3b+a+2b\right)=\frac{1}{2}\left(a+5b\right)\)
\(\sqrt{3a\left(b+2a\right)}\le\frac{1}{2}\left(5a+b\right)\)
=> \(P\le\frac{1}{2}\left(a^2+b^2+10ab\right)\)
Mà \(ab\le\frac{1}{2}\left(a^2+b^2\right)\le\frac{1}{2}.2=1\)
=> \(P\le\frac{1}{2}\left(2+10\right)=6\)
Vậy MaxP=6 khi a=b=1
\(A=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\)\(\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\)\(\left(\frac{-\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right)\)
\(=\left(\sqrt{a}+1\right)\left(-\sqrt{a}-1\right)\)
\(=-\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)=-\left(\sqrt{a}+1\right)^2\)
\(b,A=-a^2\Rightarrow-\left(\sqrt{a}+1\right)^2=a^2\)
\(\Leftrightarrow a=\sqrt{a}+1\Rightarrow a-\sqrt{a}-1=0\)
\(\Rightarrow4a-4\sqrt{a}-4=0\)
\(\Rightarrow4a-4\sqrt{a}+1-5=0\)
\(\Rightarrow\left(2\sqrt{a}-1\right)^2-\sqrt{5}^2=0\)
\(\Rightarrow\left(2\sqrt{a}-1+\sqrt{5}\right)\left(2\sqrt{a}-1-\sqrt{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{a}=1-\sqrt{5}\\2\sqrt{a}=1+\sqrt{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=\frac{1-\sqrt{5}}{2}\\\sqrt{a}=\frac{1+\sqrt{5}}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a=\frac{\left(1-\sqrt{5}\right)^2}{4}\left(tm\right)\\a=\frac{\left(1+\sqrt{5}\right)^2}{4}\left(tm\right)\end{cases}}\)
\(S=\frac{1}{B}+A=\frac{x+7}{\sqrt{x}}+\frac{\sqrt{x}+3}{\sqrt{x}}=\frac{x+\sqrt{x}+10}{\sqrt{x}}=\sqrt{x}+1+\frac{10}{\sqrt{x}}\)
\(=\sqrt{x}+\frac{10}{\sqrt{x}}+1\ge2\sqrt{\sqrt{x}.\frac{10}{\sqrt{x}}}+1=2\sqrt{10}+1\)
Dấu \(=\)khi \(\sqrt{x}=\frac{10}{\sqrt{x}}\Leftrightarrow x=10\).
\(a,x=7-4\sqrt{3}=4-2.2\sqrt{3}+3\) (Thỏa mãn ĐKXĐ)
\(=\left(2-\sqrt{3}\right)^2\)
\(B=\frac{2}{\sqrt{x}-2}=\frac{2}{\sqrt{\left(2-\sqrt{3}\right)^2}-2}\)
\(=\frac{2}{2-\sqrt{3}-2}=-\frac{2\sqrt{3}}{3}\)
\(b,P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{2}{\sqrt{x}-2}:\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}:\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(P=\frac{4}{3}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{4}{3}\)
\(\Leftrightarrow3\left(\sqrt{x}+2\right)=4\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow3\sqrt{x}+6=4\sqrt{x}+4\)
\(\Leftrightarrow6-4=4\sqrt{x}-3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(ko thỏa mãn ĐKXĐ)
=>pt vo nghiệm
d,\(\left(\sqrt{x}+1\right)P-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\frac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow\sqrt{x}+2-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)
\(\Leftrightarrow-4\sqrt{x-1}+28=-6x+10\sqrt{5x}\)
\(\Leftrightarrow x=5\)
Theo giả thiết, ta có: \(2b-ab-4\ge0\Rightarrow2b\ge ab+4\ge4\sqrt{ab}\)
\(\Rightarrow\frac{b}{\sqrt{ab}}\ge2\Rightarrow\frac{b}{a}\ge4\)
Xét \(\frac{1}{T}=\frac{ab}{a^2+2b^2}=\frac{1}{\frac{a}{b}+\frac{2b}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{16a}+\frac{31b}{16a}}\le\frac{1}{2\sqrt{\frac{1}{16}}+\frac{31}{16}.4}=\frac{4}{33}\)
\(\Rightarrow T\ge\frac{33}{4}\)
Đẳng thức xảy ra khi a = 1; b = 4
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
Lời giải:
Ta có:
$P^2=2+2(a+b)+2\sqrt{(1+2a)(1+2b)}=2+2+2\sqrt{1+2(a+b)+4ab}$
$=4+2\sqrt{3+4ab}$
Vì $a,b\geq 0$ nên $\sqrt{3+4ab}\geq \sqrt{3}$
$\Rightarrow P^2\geq 4+2\sqrt{3}$
$\Rightarrow P\geq \sqrt{3}+1$
Vậy $P_{\min}=\sqrt{3}+1$. Giá trị này được khi $(a,b)=(1,0)$ và hoán vị.