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\(\dfrac{3+\sqrt{21}}{1+\sqrt{7}}=\dfrac{\left(3+\sqrt{21}\right)\left(1-\sqrt{7}\right)}{-6}=\dfrac{3-3\sqrt{7}+\sqrt{21}-\sqrt{147}}{-6}\\ =\dfrac{3-3\sqrt{7}+\sqrt{21}-7\sqrt{3}}{-6}=-\dfrac{3-3\sqrt{7}+\sqrt{21}-7\sqrt{3}}{6}\)
ĐKXĐ: x>0
\(\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=1\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
ĐẶT x = \(\sqrt{3}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}+\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
\(\Leftrightarrow\dfrac{x}{1-\sqrt{x+1}}+\dfrac{x}{1+\sqrt{x+1}}\)
\(\Leftrightarrow\dfrac{x+x\sqrt{x+1}+x-x\sqrt{x+1}}{\left(1-\sqrt{x+1}\right).\left(1+\sqrt{x+1}\right)}\)
\(\Leftrightarrow\dfrac{2x}{1-x-1}\)
\(\Leftrightarrow\dfrac{2x}{-x}\) = -2
Mình mới làm quen toán 9, có gì sai sót mong bạn thông cảm. Chúc bạn học tốt :))
1, đk: \(x>0\) và \(x\ne4\)
Ta có: A=\(\dfrac{1}{2\sqrt{x}-x}=\dfrac{1}{-\left(x-2\sqrt{x}+1\right)+1}=\dfrac{1}{-\left(\sqrt{x}-1\right)^2+1}\)
Ta luôn có: \(-\left(\sqrt{x}-1\right)^2\le0\) với \(x>0\) và \(x\ne4\)
\(\Rightarrow-\left(\sqrt{x}-1\right)^2+1\le1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra <=> x=1 (t/m)
Vậy MinA=1 khi x=1
2, đk: \(x\ge0;x\ne1;x\ne9\)
Ta có: B=\(\dfrac{1}{x-4\sqrt{x}+3}=\dfrac{1}{\left(x-4\sqrt{x}+4\right)-1}=\dfrac{1}{\left(\sqrt{x}-2\right)^2-1}\)
Ta luôn có: \(\left(\sqrt{x}-2\right)^2\ge0\) với \(x\ge0;x\ne1;x\ne9\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2-1\ge-1\)
\(\Rightarrow B\le-1\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MaxB=-1 khi x=4
3, đk: \(x\ge0;x\ne15+4\sqrt{11}\)
Ta có: C=\(\dfrac{1}{4\sqrt{x}-x+7}=\dfrac{1}{-\left(x-4\sqrt{x}+4\right)+11}=\dfrac{1}{-\left(\sqrt{x}-2\right)^2+11}\)
Ta luôn có: \(-\left(\sqrt{x}-2\right)^2\le0\) với \(x\ge0;x\ne15+4\sqrt{11}\)
\(\Rightarrow-\left(\sqrt{x}-2\right)^2+11\le11\)
\(\Rightarrow C\ge\dfrac{1}{11}\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MinC=\(\dfrac{1}{11}\) khi x=4
\(=\sqrt{2}\left(\dfrac{2+\sqrt{5}}{2+\sqrt{5}+1}+\dfrac{2-\sqrt{5}}{2-\sqrt{5}+1}\right)\)
\(=\sqrt{2}\left(\dfrac{\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}\right)\)
\(=\sqrt{2}\cdot\dfrac{6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5}{4}\)
\(=\sqrt{2}\cdot\dfrac{2}{4}=\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(P=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(P=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)
\(P=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)
\(P=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(P=-1+\sqrt{100}=-1+10=9\)
Áp dụng:\(\dfrac{1}{\sqrt{a}+\sqrt{a+1}}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{a+1-a}=\sqrt{a+1}-\sqrt{a}\)