Bài 7:
1: Thay x=9 vào B, ta được:
\(B=\frac{1}{\sqrt9-1}=\frac{1}{3-1}=\frac12\)
2: \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
C=A:B
\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}}\)
3: C=3
=>\(x+2=3\sqrt{x}\)
=>\(x-3\sqrt{x}+2=0\)
=>\(\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}-2=0\\ \sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=2\\ \sqrt{x}=1\end{array}\right.=>\left[\begin{array}{l}x=4\left(nhận\right)\\ x=1\left(loại\right)\end{array}\right.\)
4: B<0
=>\(\frac{1}{\sqrt{x}-1}<0\)
=>\(\sqrt{x}-1<0\)
=>\(\sqrt{x}<1\)
=>0<x<1
5: \(C=\frac{x+2}{\sqrt{x}}\)
=>\(C=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\cdot\sqrt{\sqrt{x}\cdot\frac{2}{\sqrt{x}}}=2\sqrt2>2\forall x\) thỏa mãn ĐKXĐ
6: Để C nguyên thì 2⋮\(\sqrt{x}\)
=>\(\sqrt{x}\in\left\lbrace1;2\right\rbrace\)
=>x∈{1;4}
mà x<>1
nên x=4
7: \(C\ge2\sqrt2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}=\frac{2}{\sqrt{x}}\)
=>x=2
Bài 6:ĐKXĐ: x>=0; x<>25
1: \(A=\left(\frac{7}{\sqrt{x}+5}+\frac{3}{\sqrt{x}-5}-\frac{6\sqrt{x}}{x-25}\right)\cdot\frac{\sqrt{x}+5}{\sqrt{x}+8}\)
\(=\frac{7\left(\sqrt{x}-5\right)+3\left(\sqrt{x}+5\right)-6\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}+5}{\sqrt{x}+8}\)
\(=\frac{7\sqrt{x}-35+3\sqrt{x}+15-6\sqrt{x}}{\sqrt{x}-5}\cdot\frac{1}{\sqrt{x}+8}=\frac{4\sqrt{x}-20}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+8\right)}\)
\(=\frac{4}{\sqrt{x}+8}\)
2: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào A, ta được:
\(A=\frac{4}{\sqrt{\left(\sqrt5-1\right)^2}+8}\)
\(=\frac{4}{\sqrt5-1+8}=\frac{4}{\sqrt5+4}=\frac{4\left(4-\sqrt5\right)}{\left(4+\sqrt5\right)\left(4-\sqrt5\right)}\)
\(=\frac{4\left(4-\sqrt5\right)}{16-5}=\frac{4}{11}\left(4-\sqrt5\right)\)
3: \(A-\frac32=\frac{4}{\sqrt{x}+8}-\frac32\)
\(=\frac{8-3\left(\sqrt{x}+8\right)}{2\left(\sqrt{x}+8\right)}=\frac{-3\sqrt{x}-16}{2\left(\sqrt{x}+8\right)}<0\)
=>A<3/2
