cho \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}\)
a, Tìm đkxđ của A
b, Tính A khi x=\(3+2\sqrt{2}\)
c, Tìm x khi A=\(\dfrac{1}{2}\)
d,Tìm x khi A>2
e, Tìm \(x\in Z\) để A nguyên
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Lời giải:
a.
\(A=\left[\frac{(2+x)^2}{(2-x)(2+x)}+\frac{4x^2}{(2-x)(2+x)}-\frac{(2-x)^2}{(2-x)(2+x)}\right]:\frac{x(x-3)}{x^2(2-x)}\)
\(=\frac{(2+x)^2+4x^2-(2-x)^2}{(2-x)(2+x)}.\frac{x^2(2-x)}{x(x-3)}=\frac{4x(x+2)}{(2-x)(2+x)}.\frac{x^2(2-x)}{x(x-3)}=\frac{4x^2}{x-3}\)
b.
Khi $x=12$ thì $A=\frac{4.12^2}{12-3}=64$
c.
$A=1\Leftrightarrow \frac{4x^2}{x-3}=1$
$\Leftrightarrow 4x^2=x-3$
$\Leftrightarrow 4x^2-x+3=0$
$\Leftrightarrow (2x-\frac{1}{4})^2=-\frac{47}{16}< 0$ (vô lý)
Vậy không tồn tại $x$
d. Để $A$ nguyên thì $\frac{4x^2}{x-3}$ nguyên
$\Leftrightarrow 4x^2\vdots x-3$
$\Leftrightarrow 4(x^2-9)+36\vdots x-3$
$\Leftrightarrow 36\vdots x-3$
$\Leftrightarrow x-3\in\left\{\pm 1;\pm 2;\pm 3;\pm 4;\pm 9; \pm 12; \pm 36\right\}$
Đến đây bạn có thể tự tìm $x$ được rồi, chú ý ĐKXĐ để loại ra những giá trị không thỏa mãn.
e.
$A>4\Leftrightarrow \frac{4x^2}{x-3}>4$
$\Leftrightarrow \frac{x^2}{x-3}>1$
$\Leftrightarrow \frac{x^2-x+3}{x-3}>0$
$\Leftrightarrow x-3>0$ (do $x^2-x+3>0$ với mọi $x$ thuộc ĐKXĐ)
$\Leftrightarrow x>3$. Kết hợp với đkxđ suy ra $x>3$
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)
b: Thay x=-4 vào A, ta được:
\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên
1, ĐKXĐ: x\(\ge0\);x\(\ne1\)
Rút gọn P với \(x\ge0;x\ne1\)ta có
P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\dfrac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:
P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)
=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)
=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)
\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{2}+1\right)-1}{\sqrt{2}+1-2}=\dfrac{2\sqrt{2}+1}{\sqrt{2}-1}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)
\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)
\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)
e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Do \(x\ge0\)
\(\Rightarrow x\in\left\{1;9;25\right\}\)