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\(=\dfrac{8-x}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}\cdot\dfrac{\sqrt[3]{x^2}-1}{\sqrt[3]{x}\left(\sqrt[3]{x}+1\right)}\)
\(=2-\sqrt[3]{x}+\dfrac{\sqrt[3]{x}-1}{\sqrt[3]{x}-2}\)
\(=\dfrac{4-4\sqrt[3]{x}+\sqrt[3]{x^2}-\sqrt[3]{x}+1}{2-\sqrt[3]{x}}\)
\(=\dfrac{\sqrt[3]{x^2}-5\sqrt[3]{x}+5}{2-\sqrt[3]{x}}\)
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
Mẫu thức chung là (√x+1)(√x−4)
Bạn quy đồng lên rồi tính là ra
P/s: mình hơi lười. Bạn thông cảm nhé
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)Đkxđ : x>2
=(\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) \(:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)-\left(x-4\right)}\)
\(=\frac{1}{\sqrt{x}}.\frac{\sqrt{x}-2}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(A=\frac{\left(x^4+2\right)^2-x^4}{x^4+x^2+2}=\frac{\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)}{x^4+x^2+2}=x^4-x^2+2\)
\(B=\frac{a+9b+6\sqrt{ab}-4\sqrt{ab}}{\sqrt{a}+3\sqrt{b}-2\sqrt{\sqrt{ab}}}-2\sqrt{b}=\frac{\left(\sqrt{a}+3\sqrt{b}\right)^2-\left(2\sqrt{\sqrt{ab}}\right)^2}{\sqrt{a}+3\sqrt{b}-2\sqrt{\sqrt{ab}}}-2\sqrt{b}\)
\(=\frac{\left(\sqrt{a}+3\sqrt{b}-2\sqrt{\sqrt{ab}}\right)\left(\sqrt{a}+3\sqrt{b}+2\sqrt{\sqrt{ab}}\right)}{\sqrt{a}+3\sqrt{b}-2\sqrt{\sqrt{ab}}}-2\sqrt{b}\)
\(=\sqrt{a}+3\sqrt{b}+2\sqrt{\sqrt{ab}}-2\sqrt{b}\)
\(=\sqrt{a}+\sqrt{b}+2\sqrt{\sqrt{ab}}\)
\(=\left(\sqrt{\sqrt{a}}+\sqrt{\sqrt{b}}\right)^2=\left(\sqrt[4]{a}+\sqrt[4]{b}\right)^2\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)
\(\sqrt{3}+\sqrt{8-2\sqrt{15}}\\ =\sqrt{3}+\sqrt{5-2\sqrt{5\cdot3}+3}\\ =\sqrt{3}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\sqrt{3}+\sqrt{5}-\sqrt{3}=\sqrt{5}\)
\(\sqrt{x-1-2\sqrt{x-2}}\left(x\ge2\right)\\ =\sqrt{x-2-2\sqrt{x-2}+1}\\ =\sqrt{\left(\sqrt{x-2}-1\right)^2}\\ =\left|\sqrt{x-2}-1\right|\\ =\left[{}\begin{matrix}\sqrt{x-2}-1\left(\sqrt{x-2}\ge1\Leftrightarrow x\ge3\right)\\1-\sqrt{x-2}\left(\sqrt{x-2}< 1\Leftrightarrow2\le x< 3\right)\end{matrix}\right.\)
Chúc bạn học tốt nha
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