Bạn cần giúp nhanh nhưng lại không ghi đầy đủ đề bài?

a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)

\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)

\(=3\sqrt{3}\)

c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)

\(=\sqrt{7}+1-\sqrt{7}+2\)

=3

Bài 1: Ta có:

\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}}(\sqrt{6}+\sqrt{2})\)

\(=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{6-2}}(\sqrt{6}+\sqrt{2})\)

\(=\frac{\sqrt{6+2-2\sqrt{6.2}}}{2}(\sqrt{6}+\sqrt{2})\)

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

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

Bài 2:

\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)

\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)

\(=16+2\sqrt{64-4(10+2\sqrt{5})}\)

\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)

\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}\)

\(=16+2(\sqrt{20}-2)=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)

\(\Rightarrow A=\sqrt{10}+\sqrt{2}\)

Chứng minh:\(\left(\dfrac{x\sqrt{x}-3\sqrt{3x}}{x-27}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)= 1

Biến đổi vế trái ta được:

VT=\(\left(\dfrac{(x\sqrt{x}-3\sqrt{3x)}\times\left(3\sqrt{3x}+x\sqrt{x}\right)}{(x-27)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times(x^3-x^2+x)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

=\(\left(\dfrac{x^3-27x^2}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times\left(x^3-x^2+x\right)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

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

=\(\dfrac{x^3+x}{3\sqrt{3x}+x\sqrt{x}}\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)

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

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

=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{\sqrt{x}\times\left(x+3\sqrt{3}\right)}\)

=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{x\times\left(\sqrt{x}+3\sqrt{3}\right)}\)= 1 =VP

Vậy đẳng thức được chứng minh

\(a.P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\) \(b.P=a+1\)

⇔ \(2\sqrt{a}+4=a+1\)

⇔ \(a-2\sqrt{a}-3=0\)

⇔ \(a+\sqrt{a}-3\sqrt{a}-3=0\)

⇔ \(\sqrt{a}\left(\sqrt{a}+1\right)-3\left(\sqrt{a}+1\right)=0\)

⇔ \(a=9\left(TM\right)\)

KL.............

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}+\dfrac{5}{2\sqrt{2}-3}-\dfrac{5}{\sqrt{3}+\sqrt{8}}\)

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

\(=6+2\sqrt{3}\)

\(=\sqrt{3+2\sqrt{2}+1}+\sqrt{3-2\sqrt{2}+1}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}\\ =\left|\sqrt[]{3}+1\right|+\left|\sqrt{3}-1\right|-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{5}-\dfrac{5\left(\sqrt{3}-2\sqrt[]{2}\right)}{5}\\ =\sqrt{3}+1+\sqrt{3}-1-\sqrt{3}-2\sqrt{2}-\sqrt[]{3}+2\sqrt{2}\\ =0\)

a. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+14\sqrt{2}=14-14\sqrt{2}+7+14\sqrt{2}=21\)

b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}-\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)

c. \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)

b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)

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

\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)

a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)

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

c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)

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

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

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

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

b: \(=\dfrac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}+\sqrt{41}+2}}\)

\(=\dfrac{8\sqrt{41}}{\sqrt{47+5\sqrt{41}}}\)