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Chứng minh gì vậy bạn

\(\frac{B}{\sqrt{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}\)

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

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

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

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

\(B=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)

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Vài chỗ mình làm vắn tắt không hiểu cứ hỏi nhé, còn kết quả mình ấn máy tính ra chính xác rùi :)

biểu thức\(=\frac{2+\frac{2}{9}}{1-\frac{8}{9}}=\frac{\frac{20}{9}}{\frac{1}{9}}=20\)

bt = \(\frac{2+\frac{2}{9}}{\frac{4\sqrt{2}}{3}}=\frac{\frac{20}{9}}{\frac{4\sqrt{2}}{3}}=\frac{5\sqrt{2}}{6}\)

còn sai đâu ko? 

bấm máy ra kq khác, soi coi còn gì sai ko?!

Bài làm:

a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)

\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)

\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)

\(A=4+2\sqrt{3}+5\sqrt{3}-1\)

\(A=3+7\sqrt{3}\)

b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)

\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)

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

\(A=2\)

Phần b mình viết nhầm tên thành A, bn sửa thành B nhé

c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

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

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

\(C=-3\)

Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)

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

\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\frac{7-3}{2}=2\)

Vậy \(P=2\)

\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)

\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)

\(\Leftrightarrow C=-3\)

a=căn (x)

A=[(4-a^2)(2-a)+2a(4-a^2)-4a^2(2-a)]/[(4-a^2)(2-a)2a]

A=(8-10a^2+4a+3a^3)/a(16-4a^2-8a+2a^3)

A=(a-2)^2(3a+2)/a(a+2)(a-2)^2*2

A=(3a+2)/a(a+2)*2

B=2+căn(3)

A=B suy ra

(3a+2)/a(a+2)*2=2+căn 3

<=>bấm máy tính ra nghiệm a=0.1539181357

=>x=a^2 =0.02341454985

tl đúng