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a,Ta có \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)
\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)
b, Với \(x\ge0;x\ne1\)
\(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)
\(=\left(\frac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=1\)
Vậy biểu thức ko phụ thuộc biến x
c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên
thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| \(\sqrt{x}-1\) | 1 | -1 | 2 | -2 |
| \(\sqrt{x}\) | 2 | 0 | 3 | -1 |
| x | 4 | 0 | 9 | vô lí |
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
a: Thay x=9 vào A, ta được:
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)
\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)
b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)
\(\Leftrightarrow x-4=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-6=0\)
=>x=9
a) Thay x=4 vào biểu thức \(B=\dfrac{3}{\sqrt{x}-1}\), ta được:
\(B=\dfrac{3}{\sqrt{4}-1}=\dfrac{3}{2-1}=3\)
Vậy: Khi x=4 thì B=3
b) Ta có: P=A-B
\(\Leftrightarrow P=\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
a: \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
Khi x=16 thì \(B=\dfrac{2\cdot4+2}{\left(4-2\right)\left(4+2\right)}=\dfrac{10}{2\cdot6}=\dfrac{10}{12}=\dfrac{5}{6}\)
b: P=B/A
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{2}{\sqrt{x}+2}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
c: P<1
=>P-1<0
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;1;2;3\right\}\)
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 \(x=0\)
\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)
\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)
\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\) Pt vô nghiệm
Vậy không có giá trị x thỏa yêu cầu đề bài.
a: Sửa đề: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+6}\)
Khi x=4 thì \(A=\dfrac{\sqrt{4}}{\sqrt{4}+6}=\dfrac{2}{2+6}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(B=\dfrac{4}{x-1}+\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{5}{1-\sqrt{x}}\)
\(=\dfrac{4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}\)
\(=\dfrac{4+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4+x+2\sqrt{x}-3+5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+6}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Để P<0 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
mà \(\sqrt{x}>0\)
nên \(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<=x<1
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)
\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)
Vậy với x = 4 thì A = 3/4
b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}+1}\)( đpcm )