Tìm biểu thức để B = ($\frac{2x+1}{\sqrt{x^{3}}-1}$ – $\frac{\sqrt{x}}{x+\sqrt{x}+1}$)($\frac{{\sqrt{x^{3}}+1}}{{\sqrt{x}}+1}$-$\sqrt{x}$ ) với x>0 và x $\neq$ 1
a/ Rút gọn B
b/ Tìm x để B= $\frac{1}{3}$
c/ Tính B với x = 3 + $2\sqrt{2}$
Tìm biểu thức để B = ($\frac{2x+1}{\sqrt{x^{3}}-1}$ – $\frac{\sqrt{x}}{x+\sqrt{x}+1}$)($\frac{{\sqrt{x^{3}}+1}}{{\sqrt{x}}+1}$-$\sqrt{x}$ ) với
By Aubrey
Đáp án:
c) \(B = \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \left[ {\dfrac{{2x + 1 – \sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}{{\sqrt x + 1}} – \sqrt x } \right]\\
= \dfrac{{2x + 1 – x + \sqrt x }}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.\left( {x – 2\sqrt x + 1} \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\left( {\sqrt x – 1} \right)\left( {x + \sqrt x + 1} \right)}}.{\left( {\sqrt x – 1} \right)^2}\\
= \sqrt x – 1\\
b)B = \dfrac{1}{3}\\
\to \sqrt x – 1 = \dfrac{1}{3}\\
\to \sqrt x = \dfrac{4}{3}\\
\to x = \dfrac{{16}}{9}\\
c)Thay:x = 3 + 2\sqrt 2 = 2 + 2\sqrt 2 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to B = \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} – 1 = \sqrt 2 + 1 – 1 = \sqrt 2
\end{array}\)
$\begin{array}{l}\quad B =\left(\dfrac{2x +1}{\sqrt{x^3 -1}} – \dfrac{\sqrt x}{x + \sqrt x +1}\right)\left(\dfrac{\sqrt{x^3}+1}{\sqrt x+1} -\sqrt x\right)\qquad (x>0;\, x \ne 1)\\ a) \quad B = \left[\dfrac{2x +1}{(\sqrt x-1)(x + \sqrt x+1)} – \dfrac{\sqrt x(\sqrt x-1)}{(\sqrt x-1)(x + \sqrt x+1)}\right]\left[\dfrac{{(\sqrt x+1)(x – \sqrt x+1)}}{\sqrt x+1} -\sqrt x\right]\\ \to B = \dfrac{2x + 1 -x + \sqrt x}{(\sqrt x-1)(x + \sqrt x+1)}\cdot(x – \sqrt x + 1 -\sqrt x)\\ \to B = \dfrac{x + \sqrt x + 1}{(\sqrt x-1)(x + \sqrt x+1)}\cdot(x – 2\sqrt x+1)\\ \to B = \dfrac{(\sqrt x-1)^2}{\sqrt x-1}\\ \to B = \sqrt x – 1\\ b)\quad B = \dfrac13\\ \to \sqrt x – 1 = \dfrac13\\ \to \sqrt x = \dfrac43\\ \to x = \dfrac{16}{9}\\ c)\quad \text{Ta có:}\\ \quad x = 3 + 2\sqrt2\\ \to x = 2 + 2\sqrt 2 + 1\\ \to x = (\sqrt2 + 1)^2\\ \to \sqrt x = \sqrt2 + 1\\ \text{Khi đó:}\\ \quad B = \sqrt x – 1 = \sqrt2 + 1 – 1 = \sqrt2 \end{array}$