Gọi \(A=\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)

vậy bình phương

 \(A^2=4-\sqrt{7}+2.\sqrt{4-\sqrt{7}.4+\sqrt{7}}+4+\sqrt{7}\)

\(A^2=8+2\sqrt{16}=8+8=16\)

Do vậy dễ thấy \(A=\sqrt{16}=4\)

a, \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+4x}{4-x^2}\)

\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x+x+2+x^2-2x-x+2-x^2-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)

b, Thay x = 4 ta có : 

\(\frac{4^2}{\left(4-2\right)\left(4+2\right)}=\frac{16}{2.8}=\frac{16}{16}=1\)

Vậy \(A=1\)

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

bình phương 2 vế ta được : 

\(2-x=9-6\sqrt{3x+1}+3x+1\)

\(2-x-9-3x-1=-6\sqrt{3x+1}\)

\(-8-4x=-6\sqrt{3x+1}\)

\(8+4x=6\sqrt{3x+1}\)bình phương 2 vế ta được : 

\(64+64x+16x^2=108x+36\)

\(28-44x+16x^2=0\) 

PT <=> \(4\left(x-1\right)\left(4x-7\right)=0\)

\(x=1;\frac{7}{4}\)

\(P=\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}:\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(P=\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}:\left(\frac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(P=\frac{2}{\sqrt{x}-1}.\frac{x-3\sqrt{x}+2}{-5}\)

\(P=\frac{2x-6\sqrt{x}+4}{-5\sqrt{x}+5}\)

P/S: bạn tự làm nốt nha, mik chỉ rút gọn thôi : ) và bạn  kt luôn júp mik nhé : )

Đặt \(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)

\(A^2=\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)^2\)

\(A^2=\left(\sqrt{24+16\sqrt{2}}\right)^2-2\sqrt{24+16\sqrt{2}}\cdot\sqrt{24-16\sqrt{2}}+\left(\sqrt{24-16\sqrt{2}}\right)^2\)

\(A^2=\left|24+16\sqrt{2}\right|-2\sqrt{\left(24+16\sqrt{2}\right)\left(24-16\sqrt{2}\right)}+\left|24-16\sqrt{2}\right|\)

\(A^2=24+16\sqrt{2}-2\sqrt{24^2-\left(16\sqrt{2}\right)^2}+24-16\sqrt{2}\)

\(A^2=48-2\sqrt{576-512}\)

\(A^2=48-2\sqrt{64}\)

\(A^2=48-2\sqrt{8^2}=48-2\cdot\left|8\right|=32\)

=> \(A=\sqrt{32}\)

\(=\sqrt{24+12\sqrt{6}+9}\)

\(=\sqrt{(2\sqrt{6})^2+2.2\sqrt{6}.3+3^2}\)

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

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

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