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a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)
b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
Lời giải:
a. ĐKXĐ: $x>1$
\(B=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{(\sqrt{x+1}+\sqrt{x-1})^2}{2}=x+\sqrt{x^2-1}\)
b.
\(B=\frac{a^2+b^2}{2ab}+\sqrt{\frac{a^2+2ab+b^2}{2ab}.\frac{a^2-2ab+b^2}{2ab}}\)
\(=\frac{a^2+b^2}{2ab}+\sqrt{\frac{(a+b)^2(a-b)^2}{(2ab)^2}}=\frac{a^2+b^2}{2ab}+\frac{|a-b||a+b|}{|2ab|}=\frac{a^2+b^2}{2ab}+\frac{a^2-b^2}{2ab}=\frac{a}{b}\)
c.
$B\leq 1\Leftrightarrow (x-1)+\sqrt{x^2-1}\leq 0$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-1}+\sqrt{x+1})\leq 0$
$\Leftrightarrow \sqrt{x-1}\leq 0$
Mà $\sqrt{x-1}>0$ với mọi $x<1$ nên điều này vô lý)
Vậy không tồn tại $x$ thỏa đkđb
d.
$B=2\Leftrightarrow x+\sqrt{x^2-1}=2$
$\Leftrightarrow \sqrt{x^2-1}=2-x$
\(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ x^2-1=(2-x)^2=x^2-4x+4\end{matrix}\right.\)
\(\Rightarrow x=\frac{5}{4}\)
Thử lại thấy thỏa mãn
Vậy......
a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)
Xét: \(\dfrac{2\left(a+b\right)}{\sqrt{a^3}-2\sqrt{2b^3}}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{2\left(a+b\right)}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{a+2b+\sqrt{2ab}}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}=\dfrac{1}{\sqrt{a}-\sqrt{2b}}\)\(\dfrac{\sqrt{a^3}+2\sqrt{2b^3}}{2b+\sqrt{2ab}}-\sqrt{a}=\dfrac{\left(\sqrt{a}+\sqrt{2b}\right)\left(a-\sqrt{2ab}+2b\right)}{\sqrt{2b}\left(\sqrt{a}+\sqrt{2b}\right)}-\sqrt{a}=\dfrac{\left(\sqrt{a}-\sqrt{2b}\right)^2}{\sqrt{2b}}\)\(\Rightarrow P=\dfrac{\sqrt{a}-\sqrt{2b}}{\sqrt{2b}}=\sqrt{\dfrac{a}{2b}}-1\)
b, Tự lm nhé.
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)
c: Ta có: P=AB
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}+1}\)
a,ĐK:\(a>0;b>0;a\ne b\)
b,\(A=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\\ A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\\ A=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=0\)
Vậy khi A có nghĩa thì A không phụ thuộc vào a
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) Thay x=0 vào A, ta được:
\(A=\dfrac{15\cdot\sqrt{0}-11}{0+2\sqrt{0}-3}-\dfrac{3\sqrt{0}-2}{\sqrt{0}-1}-\dfrac{2\sqrt{0}+3}{\sqrt{0}+3}\)
\(=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}\)
\(=\dfrac{11}{3}-2-1\)
\(=\dfrac{11}{3}-\dfrac{9}{3}=\dfrac{2}{3}\)
`a)` A có nghĩa khi \(\left\{{}\begin{matrix}2ab\ge0\\\sqrt{a^2+b^2}\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a,b\ge0;a,b\le0\\a;b\ne0\end{matrix}\right.\)
\(\rightarrow\left[{}\begin{matrix}a;b>0\\a;b< 0\end{matrix}\right.\)
`b)`\(a=b\)
\(A=\left(1-\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)\left(1+\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)\)
\(A=1^2-\left(\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)^2\)
\(A=1-\dfrac{2ab}{a^2+b^2}\)
\(A=1-\dfrac{2a^2}{2a^2}\)
\(A=1-1=0\)