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\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)
\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)
\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\) Pt vô nghiệm
Vậy không có giá trị x thỏa yêu cầu đề bài.
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
Lời giải:
$a+b=\frac{\sqrt{6}+\sqrt{2}+\sqrt{6}-\sqrt{2}}{2}=\sqrt{6}$
$ab=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2.2}=\frac{6-2}{4}=1$
Khi đó:
$S=\frac{1}{a^7}+\frac{1}{b^7}=\frac{a^7+b^7}{a^7b^7}$
$=\frac{a^7+b^7}{(ab)^7}=\frac{a^7+b^7}{1}=a^7+b^7$
$=(a^3+b^3)(a^4+b^4)-a^3b^3(a+b)$
$=(a^3+b^3)(a^4+b^4)-(a+b)$
Ta có:
$a^3+b^3=(a+b)^3-3ab(a+b)=(\sqrt{6})^3-3\sqrt{6}=6\sqrt{6}-3\sqrt{6}=3\sqrt{6}$
$a^4+b^4=(a^2+b^2)^2-2a^2b^2=(a^2+b^2)^2-2$
$=[(a+b)^2-2ab]^2-2=(6-2)^2-2=14$
$S=3\sqrt{6}.14-\sqrt{6}=41\sqrt{6}$
a: Thay x=121 vào A, ta được:
\(A=\dfrac{121+7}{\sqrt{121}}=\dfrac{128}{11}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
c: \(S=\dfrac{1}{B}+A=\dfrac{\sqrt{x}+3}{\sqrt{x}}+\dfrac{x+7}{\sqrt{x}}=\dfrac{x+\sqrt{x}+10}{\sqrt{x}}\)
Vì \(x+\sqrt{x}+10=\sqrt{x}\left(\sqrt{x}+1\right)+10>=10>0\forall x\) thỏa mãn ĐKXĐ
và \(\sqrt{x}>0\forall\)x thỏa mãn ĐKXĐ
nên S>0 với mọi x thỏa mãn ĐKXĐ
=>S=|S|
a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)
b: Ta có: P=A:B
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a) Thay x=25 vào B ta có:
\(B=\dfrac{\sqrt{25}+2}{\sqrt{25}-2}=\dfrac{7}{3}\)
b) \(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-1}{x-5\sqrt{x}+6}\)
\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x-9-x+4+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2}{\sqrt{x}-2}\)
c) Ta có: \(A>B\) Khi:
\(\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}-2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}-\sqrt{x}< 0\\\sqrt{x}-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}-\sqrt{x}>0\\\sqrt{x}-2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow0< x< 4\)
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
