\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)

2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)

3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)

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

5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)

6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)

7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)

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

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

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

\(=\frac{12\sqrt{2}}{-2}\)

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

b) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}-\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2-\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)

\(=\frac{4\sqrt{35}}{2}\)

\(=2\sqrt{35}\)

1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

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

\(=2\sqrt{2}+\sqrt{5}+1\)

2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)

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

\(=4\sqrt{2}\)

HẾ lu van buồi

vân buồi ơi kết bạn ko

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

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

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

Ta có: 

\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)

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

\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)

Làm câu S tương tự như này rồi đối chiếu kết quả nha

Ta có: \(Q=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

\(=2\)

Sửa đề: \(S=\frac{4+\sqrt7}{3\sqrt2+\sqrt{4+\sqrt7}}+\frac{4-\sqrt7}{3\sqrt2-\sqrt{4-\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{8+2\sqrt7}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{8-2\sqrt7}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt{\left(\sqrt7+1\right)^2}}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\sqrt{\left(\sqrt7-1\right)^2}}\)

\(=\frac{\sqrt2\left(4+\sqrt7\right)}{6+\sqrt7+1}+\frac{\sqrt2\left(4-\sqrt7\right)}{6-\left(\sqrt7-1\right)}=\frac{\sqrt2\left(4+\sqrt7\right)}{7+\sqrt7}+\frac{\sqrt2\left(4-\sqrt7\right)}{7-\sqrt7}\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{2\left(4+\sqrt7\right)}{\sqrt7\left(\sqrt7+1\right)}+\frac{2\left(4-\sqrt7\right)}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{8+2\sqrt7}{\sqrt7\left(\sqrt7+1\right)}+\frac{8-2\sqrt7}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack\)

\(=\frac{1}{\sqrt2}\cdot\left\lbrack\frac{\left(\sqrt7+1\right)^2}{\sqrt7\left(\sqrt7+1\right)}+\frac{\left(\sqrt7-1\right)^2}{\sqrt7\left(\sqrt7-1\right)}\right\rbrack=\frac{1}{\sqrt2}\cdot\frac{\sqrt7+1+\sqrt7-1}{\sqrt7}=\frac{2\sqrt7}{\sqrt2\cdot\sqrt7}=\sqrt2\)

Ta có: \(R=\frac{3+\sqrt5}{2\sqrt2+\sqrt{3+\sqrt5}}+\frac{3-\sqrt5}{2\sqrt2-\sqrt{3-\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt{6+2\sqrt5}}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt{6-2\sqrt5}}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{4+\sqrt5+1}+\frac{\sqrt2\left(3-\sqrt5\right)}{4-\sqrt5+1}\)

\(=\frac{\sqrt2\left(3+\sqrt5\right)}{5+\sqrt5}+\frac{\sqrt2\left(3-\sqrt5\right)}{5-\sqrt5}=\sqrt2\cdot\left\lbrack\frac{3+\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{3-\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{6+2\sqrt5}{\sqrt5\left(\sqrt5+1\right)}+\frac{6-2\sqrt5}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\left\lbrack\frac{\left(\sqrt5+1\right)^2}{\sqrt5\left(\sqrt5+1\right)}+\frac{\left(\sqrt5-1\right)^2}{\sqrt5\left(\sqrt5-1\right)}\right\rbrack=\frac{\sqrt2}{2}\cdot\frac{\sqrt5+1+\sqrt5-1}{\sqrt5}=\frac{\sqrt2}{2}\cdot2=\sqrt2\)

Do đó: R=S

d: \(\frac{4}{\sqrt7-\sqrt3}+\frac{6}{3+\sqrt3}+\frac{\sqrt7-7}{\sqrt7-1}\)

\(=\frac{4\left(\sqrt7+\sqrt3\right)}{7-3}+\frac{6\left(3-\sqrt3\right)}{\left(3+\sqrt3\right)\left(3-\sqrt3\right)}-\frac{\sqrt7\left(\sqrt7-1\right)}{\sqrt7-1}\)

\(=\sqrt7+\sqrt3+3-\sqrt3-\sqrt7=3\)

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt5}\right)\left(4-\sqrt{10+2\sqrt5}\right)}\)

=>\(A^2=8+2\cdot\sqrt{16-10-2\sqrt5}=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)