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\(A=\left(\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(\dfrac{\sqrt{3}-1}{3\sqrt{2}-\sqrt{6}}\right)\)

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

=-4

30 tháng 9 2021

A=(√2+√3√2−√3−√2−√3√2+√3)(√3−13√2−√6)A=(2+32−3−2−32+3)(3−132−6)

=5+2√6−5+2√6−1⋅1√6=5+26−5+26−1⋅16

=-4

  
NV
13 tháng 8 2021

\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)

\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)

\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)

\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)

\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)

\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

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

Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)

\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)

\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)

\(=-11\)

3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)

\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)

\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)

\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)

12 tháng 7 2021

còn câu 1

16 tháng 7 2021

a) \(A=\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
        \(=\left(\dfrac{2}{2}-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+\dfrac{4}{2}\right)\)
        \(=\dfrac{2-\left(\sqrt{3}-1\right)}{2}:\dfrac{\left(\sqrt{3}-1\right)+4}{2}\)
        \(=\dfrac{3-\sqrt{3}}{2}.\dfrac{2}{\sqrt{3}+3}\)
        \(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}\)
        \(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
        \(=\dfrac{\left(\sqrt{3}-1\right)^2}{2}\)
Vì \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)^2>0\\2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(\sqrt{3}-1\right)^2}{2}>0\) hay A>0
=> A có căn bậc 2
Vậy......

b)\(B=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
       \(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{5}\right):\dfrac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
       \(=\left(\dfrac{\sqrt{2}\left(3-1\right)}{1-3}-\sqrt{5}\right).\dfrac{5-2}{\sqrt{5}+\sqrt{2}}\)
       \(=\left(-\sqrt{2}-\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
       \(=-\left(\sqrt{2}+\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
       \(=-3\)
Vì -3 < 0 hay B < 0 
=> B không có căn bậc 2
Vậy.....