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1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
P/s: lần sau đăng hẳn câu hỏi lên đừng có kiểu đăng như thế này, không ai muốn làm đâu
Bài này sai ngay từ đầu rồi-.-
Bài làm:
Ta có: \(x^2+\frac{1}{x^2}=7\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2\cdot x\cdot\frac{1}{x}=7\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2=7\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\)
\(\Rightarrow x+\frac{1}{x}=3\left(x>0\right)\)
Bây giờ thì dùng tam giác Pascal mà khai triển ra thôi
\(\left(x+\frac{1}{x}\right)^5=x^5+5x^4\cdot\frac{1}{x}+10x^3\cdot\frac{1}{x^2}+10x^2\cdot\frac{1}{x^3}+5x\cdot\frac{1}{x^4}+\frac{1}{x^5}\)
\(=x^5+5x^3+10x+\frac{10}{x}+\frac{5}{x^3}+\frac{1}{x^5}=\left(x^5+\frac{1}{x^5}\right)+5\left(x^3+\frac{1}{x^3}\right)+10\left(x+\frac{1}{x}\right)\)
\(\Rightarrow x^5+\frac{1}{x^5}=\left(x+\frac{1}{x}\right)^5-5\left(x^3+\frac{1}{x^3}\right)-10\left(x+\frac{1}{x}\right)\)
\(=3^5-5\left(x+\frac{1}{x}\right)\left(x^2-x\cdot\frac{1}{x}+\frac{1}{x^2}\right)-10\cdot3\)
\(=243-5\cdot3\cdot\left(7-1\right)-30=123\)
Vậy \(x^5+\frac{1}{x^5}=123\)