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a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{4\sqrt{x}}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
Bài 5:
a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:
\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)
b: Để E<1 thì E-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
c: Để E nguyên thì \(4⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)
hay \(x\in\left\{16;25;49\right\}\)
Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
Thay \(x=\sqrt{3}-1\) vào \(B\), ta được
\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)
b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)
Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)
Vậy \(B_{min}=-2\) khi \(x=0\)
\(A=\dfrac{\left(\sqrt{x}-2\right)^2+1}{\sqrt{x}-2}=\sqrt{x}-2+\dfrac{1}{\sqrt{x}-2}\\ \ge2\sqrt{\left(\sqrt{x}-2\right)\left(\dfrac{1}{\sqrt{x}-2}\right)}=2\cdot1=2\left(BĐT.cauchy\right)\)
Dấu \("="\Leftrightarrow\left(\sqrt{x}-2\right)^2=1\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
\(A=\dfrac{x-4\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)^2+1}{\sqrt{x}-2}=\sqrt{x}-2+\dfrac{1}{\sqrt{x}-2}\)
Áp dụng bất đẳng thức Cauchy cho 2 số dương:
\(A=\sqrt{x}-2+\dfrac{1}{\sqrt{x}-2}\ge2\sqrt{\dfrac{\sqrt{x}-2}{\sqrt{x}-2}}=2\)
\(minA=2\Leftrightarrow\sqrt{x}-2=1\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
Với các số thực không âm a; b ta luôn có BĐT sau:
\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) (bình phương 2 vế được \(2\sqrt{ab}\ge0\) luôn đúng)
Áp dụng:
a.
\(A\ge\sqrt{x-4+5-x}=1\)
\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(A\le\sqrt{\left(1+1\right)\left(x-4+5-x\right)}=\sqrt{2}\) (Bunhiacopxki)
\(A_{max}=\sqrt{2}\) khi \(x-4=5-x\Leftrightarrow x=\dfrac{9}{2}\)
b.
\(B\ge\sqrt{3-2x+3x+4}=\sqrt{x+7}=\sqrt{\dfrac{1}{3}\left(3x+4\right)+\dfrac{17}{3}}\ge\sqrt{\dfrac{17}{3}}=\dfrac{\sqrt{51}}{3}\)
\(B_{min}=\dfrac{\sqrt{51}}{3}\) khi \(x=-\dfrac{4}{3}\)
\(B=\sqrt{3-2x}+\sqrt{\dfrac{3}{2}}.\sqrt{2x+\dfrac{8}{3}}\le\sqrt{\left(1+\dfrac{3}{2}\right)\left(3-2x+2x+\dfrac{8}{3}\right)}=\dfrac{\sqrt{510}}{6}\)
\(B_{max}=\dfrac{\sqrt{510}}{6}\) khi \(x=\dfrac{11}{30}\)
a)Ta có:A=\(\sqrt{x-4}+\sqrt{5-x}\)
=>A2=\(x-4+2\sqrt{\left(x-4\right)\left(5-x\right)}+5-x\)
=>A2= 1+\(2\sqrt{\left(x-4\right)\left(5-x\right)}\ge1\)
=>A\(\ge\)1
Dấu '=' xảy ra <=> x=4 hoặc x=5
Vậy,Min A=1 <=>x=4 hoặc x=5
Còn câu b tương tự nhé
a: \(A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
\(y=\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}+\sqrt{\frac{x^2}{4}-\sqrt{x^2-4}}\) Điều kiện: \(x\ge2\)
\(\Rightarrow2y=2.\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}+2.\sqrt{\frac{x^2}{4}-\sqrt{x^2-4}}\)
\(=\sqrt{x^2+4\sqrt{x^2-4}}+\sqrt{x^2-4\sqrt{x^2-4}}\)
\(=\sqrt{x^2-4+4\sqrt{x^2-4}+4}+\sqrt{x^2-4-4\sqrt{x^2-4}+4}\)
\(=\sqrt{\left(\sqrt{x^2-4}+2\right)^2}+\sqrt{\left(\sqrt{x^2-4}-2\right)^2}\)
\(=\left|\sqrt{x^2-4}+2\right|+\left|\sqrt{x^2-4}-2\right|\)
\(=\sqrt{x^2-4}+2+\left|\sqrt{x^2-4}-2\right|\)(1)
TH1: \(\sqrt{x^2-4}-2\ge0\Rightarrow\sqrt{x^2-4}\ge2\Rightarrow x^2-4\ge4\Rightarrow x\ge2\sqrt{2}\).Ta có:
\(\left(1\right)=\sqrt{x^2-4}+2+\sqrt{x^2-4}-2=2\sqrt{x^2-4}\)
Do \(x\ge2\sqrt{2}\Rightarrow2\sqrt{x^2-4}\ge2\sqrt{\left(2\sqrt{2}\right)^2-4}=4\)
TH2: \(\sqrt{x^2-4}-2< 0\Rightarrow\sqrt{x^2-4}< 2\Rightarrow x^2-4< 4\Rightarrow x^2< 8\Rightarrow2\le x< 2\sqrt{2}\).Ta có:
\(\left(1\right)=\sqrt{x^2-4}+2-\sqrt{x^2-4}+2=4\)
Vậy GTNN của y bằng 4.
Dấu "=" xảy ra khi \(2\le x\le2\sqrt{2}\)