bài 1:
+) ta có : \(S=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
\(\Leftrightarrow S=\dfrac{3\sqrt{2}+\sqrt{10}}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}+\dfrac{3\sqrt{2}-\sqrt{10}}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\)
\(\Leftrightarrow S=\dfrac{3\sqrt{2}+\sqrt{10}}{3\sqrt{5}+1}+\dfrac{3\sqrt{2}-\sqrt{10}}{3\sqrt{5}-1}=\dfrac{20\sqrt{10}}{\left(3\sqrt{5}\right)^2-1^2}=\dfrac{20\sqrt{10}}{44}=\dfrac{5\sqrt{10}}{11}\)
+) ta có : \(T=\dfrac{4+\sqrt{7}}{2\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{2\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
\(\Leftrightarrow T=\dfrac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{8+2\sqrt{7}}}+\dfrac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{8-2\sqrt{7}}}\)
\(\Leftrightarrow T=\dfrac{4\sqrt{2}+\sqrt{14}}{5+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{5-\sqrt{7}}=\dfrac{54\sqrt{2}}{5^2-\left(\sqrt{7}\right)^2}=\dfrac{54\sqrt{2}}{18}=3\sqrt{2}\)
Bài 6:
Mấy bài này bạn chịu khó tra gg là có mà
bài 6 : ta có : \(x^2+y^2=a^2+b^2\Leftrightarrow\left(x+y\right)^2-2xy=\left(a+b\right)^2-2ab\)
\(\Leftrightarrow xy=ab\) kèm theo \(x+y=a+b\)
\(\Rightarrow\left\{{}\begin{matrix}x=a\\y=b\end{matrix}\right.\) \(\Leftrightarrow P=x^n+y^n=a^n+b^n\)
bài 6: ta có : \(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}=\dfrac{2006-c}{c}+\dfrac{2006-a}{a}+\dfrac{2006-b}{b}\)
\(=\dfrac{2006}{a}+\dfrac{2006}{b}+\dfrac{2006}{c}-3=2006\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-3=4022027\)
bài 10 : từ \(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\Rightarrow x=???\)
thế vào \(A\) là xong
ta có \(\sqrt{4}< \sqrt{4.\sqrt{5...\sqrt{2000}}}\)
\(\Rightarrow\sqrt{2+\sqrt{3.\sqrt{4.....\sqrt{2000}}}}>\sqrt{2+\sqrt{3.\sqrt{4}}}>2\)
\(\Rightarrow\) đề sai :))
ta có : \(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\Rightarrow x=x^2-2x+1\)
ta có : \(A=\dfrac{x^5-4x^3-3x+9}{x^4+3x^2+11}\) \(=\dfrac{3x^5-12x^3-9x+27}{3x^4+9x^2+33}\)
\(=\dfrac{\left(3x^5-6x^4+3x^3\right)+6x^4-15x^3-9x+27}{3x^4+9x^2+33}\)
\(=\dfrac{9x^4-15x^3-9x+27}{3x^4+9x^2+33}=\dfrac{2x^4+7x^4-14x^3+7x^2-x^3-7x^2-9x+27}{3x^4+9x^2+33}\)\(=\dfrac{2x^4+6x^3-12x^2+6x+5x^2-15x+27}{3x^4+9x^2+33}\)
\(=\dfrac{2x^4+6x^2+22+5x^2-10x+5-5x}{3x^4+9x^2+33}=\dfrac{2x^4+6x^2+22}{3x^4+9x^2+33}=\dfrac{2}{3}\)
Bài 8.
Ta có:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}}=\dfrac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\)
\(=\sqrt{k}\left(\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(=\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(A=\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}\)
\(\Rightarrow A< 2\left[\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+\left(\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}\right)+...+\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\right]\)
\(\Rightarrow A< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{n+1}}\right)< 2.\dfrac{1}{\sqrt{2}}=\sqrt{2}\)
Vậy \(A< \sqrt{2}\)
Dành cả thanh xuân chắc t cũng giải không hết cái đống này:))
tranxuanrin Chịu khó tra mạng với suy nghĩ thử đi bạn.
Cho t xin 1 slot bài 5.
Bài 5.
Ta có : \(10+\sqrt{24}+\sqrt{40}+\sqrt{60}=\left(2+3+5\right)+2\sqrt{2.3}+2\sqrt{2.5}+2\sqrt{3.5}\)
\(=\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
\(\Rightarrow x=1\)
Vậy x=1
Cho t xin 1 slot bài 3 luôn , cơ mà chờ tý :))
