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BT
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BT
20 tháng 7 2017 lúc 19:46

b, tìm x thuộc Z để B thuộc Z

c, Tìm x thuộc R để B có giá trị nguyên

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MN
7 tháng 3 2020 lúc 13:35

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

a) \(B=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{x-9}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)

\(\Leftrightarrow B=\left(1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{x-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\frac{x-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{3}{\sqrt{x}+3}:\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{3}{\sqrt{x}+3}:\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(\Leftrightarrow B=\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

\(\Leftrightarrow B=\frac{3}{\sqrt{x}+2}\)

b) ??

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MN
7 tháng 3 2020 lúc 13:39

b) Để \(B\inℤ\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1\right\}\)(Loại các giá trị âm)

\(\Leftrightarrow x=1\)

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HB
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KW
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TA
8 tháng 7 2017 lúc 22:54

Đặt  \(\sqrt{x}=t\)  rồi rút gọn bình thường

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H24
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TD
5 tháng 7 2019 lúc 8:42

a) \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2b}{a-b}\)

\(=\frac{a+b+2\sqrt{ab}}{2\left(a-b\right)}-\frac{a+b-2\sqrt{ab}}{2\left(a-b\right)}+\frac{4b}{2\left(a-b\right)}=\frac{a+b+2\sqrt{ab}-a-b+2\sqrt{ab}+4b}{2\left(a-b\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(a-b\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)}\)

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VT
5 tháng 7 2019 lúc 9:01

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{a-b}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)\(=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

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NA
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NT
29 tháng 1 2022 lúc 23:36

\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+9+2x-4\sqrt{x}-5}\)

\(=\dfrac{3\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}=\dfrac{3}{\sqrt{x}-2}\)

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KO
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NH
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C =  \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\)\(\left(\frac{-\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)(  \(x\ge0\) , \(x\ne9;4\))

 =  \(\frac{x-9-x+3\sqrt{x}}{x-9}\)\(\frac{9-x+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(\frac{3\sqrt{x}-9}{x-9}\)\(\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

=  \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(:\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(\frac{3}{\sqrt{x}-2}\)

#mã mã#

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HB
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ND
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CD
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NH
6 tháng 7 2017 lúc 20:07

\(\left(\frac{x+3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\frac{\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\frac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-3\cdot\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-3}{-\left(\sqrt{x}-2\right)}=\frac{3}{\sqrt{x}-2}\)

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