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bạn quy đồng nha,,nhóm cái căn3 + căn 5 thành 1 nhóm,,,rồi quy đồng \(\sqrt{2}-\left(\sqrt{3}+\sqrt{5}\right)\)

\(A=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

      \(=\frac{6+2\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)

        \(=\frac{6+2\sqrt{5}}{2+\sqrt{5+2\sqrt{5}+1}}+\frac{6-2\sqrt{5}}{2+\sqrt{5-2\sqrt{5}+1}}\)

           \(=\frac{6+2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6-2\sqrt{5}}{2+\sqrt{\left(\sqrt{5}-1\right)^2}}\)

             \(=\frac{6+2\sqrt{5}}{2+\left|\sqrt{5}+1\right|}+\frac{6-2\sqrt{5}}{2-\left|\sqrt{5}-1\right|}\)

               \(=\frac{6+2\sqrt{5}}{2+\sqrt{5}+1}+\frac{6-2\sqrt{5}}{2-\sqrt{5}+1}\)( vì \(\sqrt{5}+1>0;\sqrt{5}-1>0\))

               \(=\frac{6+2\sqrt{5}}{3+\sqrt{5}}+\frac{6-2\sqrt{5}}{3-\sqrt{5}}\)

                 \(=2+2\)

                   \(=4\)

Vậy A = 4

Tích cho mk nhoa !!!! ~~

\(A=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\)

\(=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\sqrt{3}}{3-1}\)

\(=\frac{2\sqrt{3}}{2}\)

\(=\sqrt{3}\)

\(B=\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}\)

\(=\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}+\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}\)

\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)

\(=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\frac{12}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\frac{12}{5-1}\)

\(=\frac{12}{4}\)

\(=3\)

Xét biểu thức phụ : 1 (2n+3)√2n+1+(2n+1)√2n+3 = 1 √2n+1.√2n+3(√2n+1+√2n+3) = √2n+3−√2n+1 √2n+1.√2n+3[(2n+3)−(2n+1)] = √2n+3−√2n+1 2√2n+1.√2n+3 = 1 2 ( 1 √2n+1 − 1 √2n+3 )với n≥1 Áp dụng : S= 1 3√1+1√3 + 1 3√5+5√3 + 1 5√7+7√5 +...+ 1 101√103+103√101 = 1 2 ( 1 √1 − 1 √3 )+ 1 2 ( 1 √3 − 1 √5 )+ 1 2 ( 1 √5 − 1 √7 )+...+ 1 2 ( 1 √101 − 1 √103 ) = 1 2 (1− 1 √3 + 1 √3 − 1 √5 + 1 √5 − 1 √7 +...+ 1 √101 − 1 √103 ) = 1 2 (1− 1 √103 )

Xem kỹ lại đề nhé! loại này đề lệch một tý thôi -->Không rút được !

p/s: Tránh truongf hợp làm đến cuối mới biết đề sai.

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)