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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
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
A) ĐKXĐ : \(x\ge0\) và \(x\ne4\)
Rút gọn :\(A=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{4\sqrt{x}}{4-x}\)
\(A=\frac{2\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{4-2\sqrt{x}+2+\sqrt{x}+4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{6+3\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3}{2-\sqrt{x}}\)
b) thay \(x=7+4\sqrt{3}\) vào A
ta được :\(A=\frac{3}{2-\sqrt{7+4\sqrt{3}}}=\frac{3}{2-2+\sqrt{3}}=\frac{3}{\sqrt{3}}\)
vậy vói \(x=7+4\sqrt{3}\) thì \(A=\frac{3}{\sqrt{3}}\)
c)với\(x\ge0\) và \(x\ne4\)
Để \(A=-\frac{3}{7}\Leftrightarrow\frac{3}{2-\sqrt{x}}=-\frac{3}{7}\)
\(\Leftrightarrow3.7=-3\left(2-\sqrt{x}\right)\)
\(\Leftrightarrow21=-6+3\sqrt{x}\)
\(\Leftrightarrow21+6=3\sqrt{x}\)
\(\Leftrightarrow27=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=9\)
\(\Leftrightarrow x=81\)
Vậy để\(A=-\frac{3}{7}\Leftrightarrow x=81\)
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
a, \(B=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)(ĐK: \(x\ne1\))
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
\(=\frac{x-2\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b, ĐK: \(x\ne1\)
\(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\sqrt{5}+2-\sqrt{5}+2=4\)
Thay \(x=4\left(TM\right)\)vào B ta có:
\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)
Vậy với \(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)nên \(B=\frac{1}{3}\)
c. ĐK: \(x\ne1\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\)\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le2\Leftrightarrow\frac{-2}{\sqrt{x}+1}\ge-2\)\(\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge-1\)
Dấu = xảy ra \(\Leftrightarrow\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy \(MinB=-1\Leftrightarrow x=0\)
d, ĐK: \(x\ne1\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(B\inℤ\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\inℤ\Leftrightarrow\frac{2}{\sqrt{x}+1}\inℤ\)\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\Leftrightarrow\sqrt{x}+1\in\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy với \(x=0\)thì \(B\inℤ\)
ĐKXĐ: x > 4
a, Có \(A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
\(\orbr{\begin{cases}=2\sqrt{x-4}\left(với\sqrt{x-4}\ge2\right)\\=4\left(với\sqrt{x-4}< 2\right)\end{cases}}\)
b, Xét \(A=2\sqrt{x-4}\)thì \(\sqrt{x-4}\ge2\)
\(\Leftrightarrow x-4\ge4\)
\(\Leftrightarrow x\ge8\)
Khi đó \(A=2\sqrt{x-4}\ge2\sqrt{8-4}=4\)
Nên \(A_{min}=4\Leftrightarrow x=8\)
c, Với \(x=\sqrt{15+\sqrt{6}}\)thì \(\sqrt{x-4}=\sqrt{\sqrt{15+\sqrt{6}}-4}< 2\)
Nên từ câu a => A = 4