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Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
Lời giải:
\(P=\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(\frac{x+2}{\sqrt{x^3}-1}+\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\sqrt{x^3}-1}+\frac{x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}=\frac{2x+1}{\sqrt{x^3}-1}-\frac{x+\sqrt{x}+1}{\sqrt{x^3}-1}\)
\(=\frac{2x+1-(x+\sqrt{x})}{\sqrt{x^3}-1}=\frac{x-\sqrt{x}}{\sqrt{x^3}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(P-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{2\sqrt{x}-(x+1)}{3(x+\sqrt{x}+1)}\)
\(=\frac{-(\sqrt{x}-1)^2}{3(x+\sqrt{x}+1)}\)
Với \(x\neq 1, x\geq 0\Rightarrow -(\sqrt{x}-1)^2< 0; x+\sqrt{x}+1>0\)
Do đó: \(P-\frac{1}{3}< 0\Rightarrow P< \frac{1}{3}\)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
Lời giải:
a)
Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)
Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)
\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)
b)
ĐK: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :
\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)
Vậy..........
a, \(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\left(đkxđ:x\ge0,x\ne4\right)\)
\(A=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\dfrac{2-\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\dfrac{-2\sqrt{x}+4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\dfrac{2\left(-\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)
\(=\dfrac{a}{a-b}\)
a) \(=\frac{x^2-\sqrt{3^2}}{x+\sqrt{3}}=\frac{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}=a+\sqrt{a+1}\)
a. \(\dfrac{x^2-3}{x+\sqrt{3}}=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)