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Hãy so sánh \(a^4-2a^3+a^2\) với 0
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\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\left(ĐK:x\ne\pm1\right)\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\left(\sqrt{x}-3\right)}{x-1}\)
\(=\frac{2\sqrt{x}-6}{x-1}\)
\(\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\left(\frac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\left(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-1-2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{x-2\sqrt{x}-3}{\sqrt{x}+1}\right)\)
\(=\left(\frac{2\sqrt{x}\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\)
\(=\frac{2x\sqrt{x}-6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\)
Bài lm có sai sót chỗ nào thì mong bn sửa lại........
\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x-4\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{x-3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\sqrt{x}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+3}\)
\(=\frac{x+3\sqrt{x}-x+4\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{7\sqrt{x}-3}{\sqrt{x}+3}\)
\(\frac{x-3\sqrt{x}}{\sqrt{x}-3}-\frac{x+4\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\sqrt{x}-\sqrt{x}-1=-1\)
\(a,\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}-1+\sqrt{5}+1\)
\(=2\sqrt{5}\)
\(b,\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)(Đề sai sửa lại)
\(=\sqrt{5-2.2\sqrt{5}+4}+\sqrt{5+2.2\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2+\sqrt{5}+2\)
\(=2\sqrt{5}\)
\(x-5\sqrt{x-2}=-2\)
\(\Leftrightarrow-5\sqrt{x-2}=-2-x\)
\(\Leftrightarrow\left(-5\sqrt{x-2}\right)^2=-2x-x\)
<=> 25x - 50 = 4 + 4x + x2
<=> x = 18 hoặc x = 3
Vậy:...
\(DKXĐ:x\ge2\)
\(x-5\sqrt{x-2}=-2\)
\(\Leftrightarrow5\sqrt{x-2}=x+2\)
\(\Leftrightarrow25\left(x-2\right)=\left(x+2\right)^2\)
\(\Leftrightarrow25x-50=x^2+4x+4\)
\(\Leftrightarrow x^2+4x+4-25x+50=0\)
\(\Leftrightarrow x^2-21x+54=0\)
\(\Leftrightarrow\left(x-18\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-18=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=18\\x=3\end{cases}\left(\frac{t}{m}ĐKXĐ\right)}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Rightarrow\sqrt{2x-3}=2\sqrt{x-1}\)
\(\Rightarrow2x-3=4\left(x-1\right)\)
\(\Rightarrow2x-3=4x-4\)
\(\Rightarrow4x-2x=4-3\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
\(ĐKXĐ:\hept{\begin{cases}x-1>0\\2x-3\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>1\\x>\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow x>\frac{3}{2}\)
Vậy nên \(x=\frac{1}{2}\) không thỏa mãn ĐKXĐ.
\(a^4-2a^3+a^2\)
\(=a^2\left(a^2-2a+1\right)\)
\(=a^2\left(a-1\right)^2\ge0\)
Vậy \(a^4-2a^3+a^2\ge0\)