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a) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}\ne5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Khi \(x=16\Rightarrow A=\dfrac{\sqrt[]{16}+2}{\sqrt[]{16}-5}=\dfrac{4+2}{4-5}=-6\)
b) \(B=\dfrac{3}{\sqrt[]{x}+5}+\dfrac{20-2\sqrt[]{x}}{x-25}\)
B có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{3\left(\sqrt[]{x}-5\right)+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{3\sqrt[]{x}-15+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{x}+5}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{1}{\sqrt[]{x}-5}\left(dpcm\right)\)
c) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\in Z\left(x\in Z\right)\)
\(\Leftrightarrow\sqrt[]{x}+2⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}+2-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}+2-\sqrt[]{x}+5⋮\sqrt[]{x}-5\)
\(\Leftrightarrow7⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}-5\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)
\(\Leftrightarrow x\in\left\{16;36;144\right\}\)
d) \(A>B\left(2\sqrt[]{x}+5\right)\)
\(\Leftrightarrow\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}>\dfrac{1}{\sqrt[]{x}-5}\left(2\sqrt[]{x}+5\right)\)
\(\Leftrightarrow\sqrt[]{x}+2>2\sqrt[]{x}+5\)
\(\Leftrightarrow\sqrt[]{x}< -3\)
mà \(\sqrt[]{x}\ge0\)
\(\Leftrightarrow x\in\varnothing\)
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12:
\(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\dfrac{\sqrt{y}}{y}\cdot\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
a) \(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}\)
\(=\sqrt{\dfrac{x}{y}}\)
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c: Thay x=-2 và y=-3 vào (1), ta được:
-4m+6-1=-3
=>-4m=-8
hay m=2
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Bài 3:
\(a,m=-2\Leftrightarrow y=-3x-3\\ b,\Leftrightarrow\left\{{}\begin{matrix}m-1=2\\2m+1\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=3\\m\ne0\end{matrix}\right.\Leftrightarrow m=3\\ c,\text{PT hoành độ giao điểm: }2x-7=\left(m-1\right)x+2m+1\\ \text{Thay }x=2\\ \Leftrightarrow2m-2+2m+1=-3\\ \Leftrightarrow4m=-2\Leftrightarrow m=-\dfrac{1}{2}\\ d,\text{Gọi điểm cần tìm là }A\left(x_0;y_0\right)\\ \Leftrightarrow y_0=\left(m-1\right)x_0+2m+1\\ \Leftrightarrow mx_0-x_0+2m+1-y_0=0\\ \Leftrightarrow m\left(x_0+2\right)+\left(1-x_0-y_0\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0+2=0\\1-x_0-y_0=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=-2\\y_0=3\end{matrix}\right.\Leftrightarrow A\left(-2;3\right)\)
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(p) đi qua A(-1;2)
=> 2 = (m - 2).(-1)2
<=> m - 2 = 2
<=> m = 4
Vậy m = 4 thì (p) đi qua A(-1 ; 2)
\(d,ĐKXĐ:x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)`
\(2x-3\sqrt{x}-9=0\)
\(\Delta=\sqrt{ \left(-3\right)^2-4.2.\left(-9\right)}=9\)
\(\orbr{\begin{cases}\sqrt{x}=\frac{3+9}{4}=3\left(TM\right)\\x=\frac{3-9}{4}=-\frac{3}{2}\left(KTM\right)\end{cases}}\Rightarrow x=3^2=9\left(TM\right)\)
vậy nghiệm duy nhất của pt là 9