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Mik sẽ viết lại đề bài.Bạn cs thể giải đầy đủ cho mik giùm nhen ko cần ngắn cứ dài . Cảm ơn
A=\(\sqrt{7}-4\sqrt{3}+\sqrt{4}-2\sqrt{3}\)
B=\(\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)\) \(\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
C=\(\left(\sqrt{3}+1\right)\) \(\frac{\sqrt{14}-6\sqrt{3}}{5+\sqrt{3}}\)
nguyen thao:
Câu A: vẫn giống ban đầu mà bạn? Mình nghĩ bạn vẫn viết sai đề. Đề đúng là \(A=\sqrt{7-4\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B=\left[2+\frac{\sqrt{5}(\sqrt{5}-1)}{\sqrt{5}-1}\right]\left[2-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1)}\right]\)
\(=(2+\sqrt{5})(2-\sqrt{5})=2^2-(\sqrt{5})^2=4-5=-1\)
$C=\frac{(\sqrt{3}+1)(\sqrt{14}-6\sqrt{3})}{5+\sqrt{3}}$
$=\frac{-18-6\sqrt{3}+\sqrt{14}+\sqrt{42}}{5+\sqrt{3}}$ vẫn xấu lắm bạn ạ :''>
a,\(\left(\sqrt{6}-\sqrt{10}\right)\sqrt{4+\sqrt{15}}=\sqrt{6}.\sqrt{4-\sqrt{15}}-\sqrt{10}.\sqrt{4+\sqrt{15}}\)
=\(\sqrt{24+6\sqrt{15}}-\sqrt{40+10\sqrt{15}}=\sqrt{\left(\sqrt{15}+3\right)^2}-\sqrt{\left(\sqrt{15}+5\right)^2}\)
=\(\sqrt{15}+3-\sqrt{15}-5=-2\)
b,\(\left(\sqrt{3}+\sqrt{30}\right)\sqrt{10-\sqrt{41-4\sqrt{10}}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40-2\sqrt{40}+1}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{\left(\sqrt{40}-1\right)^2}}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40}+1}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{11-2\sqrt{10}}=\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{\left(\sqrt{10}-1\right)^2}\)
=\(\sqrt{3}\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)=9\sqrt{3}\)
2,\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}\left(1-\sqrt{a}\right)-\sqrt{a}+4}{1-a}\right)\)
\(A=\left(\frac{a+\sqrt{a}-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}-a-\sqrt{a}+4}{1-a}\right)=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}\right).\left(\frac{1-a}{4-a}\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}+1}.\frac{a-1}{a-4}=\frac{\sqrt{a}-1}{\sqrt{a}+2}\)
b, ̣để \(A=\frac{1}{2}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}+2}=\frac{1}{2}\Leftrightarrow2\sqrt{a}-2=\sqrt{a}+2\Leftrightarrow\sqrt{a}=4\Leftrightarrow a=16\left(t.m\right)\)
Bạn oi bài 2 hàng A thú 2 phải là \(\frac{\sqrt{a}-2}{\sqrt{a}+1}\) mình nhầm
\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{2-\sqrt{x}}+\dfrac{3\sqrt{x}-2}{x-2}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{2\sqrt{x}-x}\right)=\dfrac{x-2\sqrt{x}+3\sqrt{x}+6+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b)\(\frac{3\sqrt{2}}{\sqrt{3}+1}\)
\(=\frac{3\sqrt{2}\left(\sqrt{3}-1\right)}{(\sqrt{3}+1)\left(\sqrt{3}-1\right)}\)
\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{3-1}\)
\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{2}\)
a)\(\left(\sqrt{3}+1\right)^2+\left(1-\sqrt{3}\right)^2\)
\(=3+2\sqrt{3}+1+1-2\sqrt{3}+3\)
\(=8\)
b)\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{28.7}-2\sqrt{3.7}+\sqrt{7}.\sqrt{7}+\sqrt{84}\)
\(=\sqrt{196}-2\sqrt{21}+7+\sqrt{4.21}\)
\(=\sqrt{14^2}-2\sqrt{21}+7+2\sqrt{21}\)
\(=14-2\sqrt{21}+7+2\sqrt{21}\)
\(=21\)