mình chỉ biết mỗi kq rút gọn thôi còn chi tiết thì mình ko rõ lắm

Ta có :\(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)

              \(=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\)

              \(=8-\frac{2}{\sqrt{2}}\sqrt{4+2\sqrt{3}}-2\sqrt{3\left(2^2-\sqrt{2+\sqrt{3}}^2\right)}\)       

              \(=8-\sqrt{2}\sqrt{\sqrt{3}^2+2\cdot1\sqrt{3}+1^2}-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)

              \(=8-\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{3}\sqrt{2-\sqrt{3}}\)

               \(=8-\sqrt{2}\left(\sqrt{3}+1\right)-\frac{2\sqrt{3}}{\sqrt{2}}\sqrt{4-2\sqrt{3}}\)

               \(=8-\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{6}\sqrt{\left(\sqrt{3}-1\right)^2}\)

               \(=8-\sqrt{6}-\sqrt{2}-\sqrt{6}\left(\sqrt{3}-1\right)\)

               \(=8-\sqrt{6}-\sqrt{2}-\sqrt{18}+\sqrt{6}\)

               \(=8-\sqrt{2}-\sqrt{18}\)

               \(=8-\sqrt{2}\left(3+1\right)=8-4\sqrt{2}\)

\(\Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16\left(8-4\sqrt{2}\right)\)

                          \(=8^2+4^2\cdot\sqrt{2}^2-2\cdot8\cdot4\sqrt{2}-16\cdot8+16\cdot4\sqrt{2}\)

                          \(=64+32-64\sqrt{2}-128+64\sqrt{2}\)

                          \(=-32\)

         Vậy \(x^4-16x^2=-32\)

Tại hạ làm bừa có gì mong đạo hữu lượng thứ =))

a) \(\sqrt{2,5.2560}=\sqrt{25.256}=\sqrt{25}.\sqrt{256}=5.16=80\)

b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{\frac{7}{2}}.\sqrt{\frac{5}{2}}.\sqrt{7}.\sqrt{\frac{1}{5}}\)

\(=\sqrt{\frac{7}{2}.\frac{5}{2}.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)

c) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1}.\sqrt{0,09}\)

\(=\sqrt{4.121}.\sqrt{9.0,01}=\sqrt{4}.\sqrt{121}.\sqrt{9}.\sqrt{0,01}\)

\(=2.11.3.0,1=6,6\)

bình phương 2 vế lên ta được 

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

\(< =>2x+2\sqrt{x^2-4x+1}=\frac{x^2+6x+9}{4}\)

\(< =>2\sqrt{x^2-4x+1}=\frac{x^2-2x+9}{4}\)

\(< =>\sqrt{x^2-4x+1}=\frac{x^2-2x+9}{8}\)

tiếp tục mình phương 2 vế thì sẽ ra

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

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

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

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

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

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

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

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

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

\(=1-3=-2\)

Xét phân số tổng quát là: 

\(A=\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}\)

=>    \(A< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}.\sqrt{n+1}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Thay từng số 1; 2; ....;  48 vào phân số tổng quát A

=>   \(S< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

=>   \(S< \frac{1}{2}\left(1-\frac{1}{7}\right)=\frac{1}{2}.\left(\frac{6}{7}\right)=\frac{3}{7}\)

VẬY    \(S< \frac{3}{7}\)

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

P = \(\left(\frac{\sqrt{a}.\sqrt{a}-1}{2\sqrt{a}}\right)^2\cdot\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

P = \(\frac{\left(a-1\right)^2}{4a}\cdot\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)

P = \(\frac{a-1}{4\sqrt{a}^2}\cdot\left(-4\sqrt{a}\right)\)

P = \(\frac{1-a}{\sqrt{a}}\)

b) với x > 0 và x khác 1

P < 0 => \(\frac{1-a}{\sqrt{a}}< 0\)

Do \(\sqrt{a}>0\) => 1 - a < 0 => a > 1

Vậy S = {a|a > 1}

Có 1 kiểu hơi khác Conan 1 tí -.-

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

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

\(=\frac{\left(1-a\right).4\sqrt{a}}{4a}=\frac{1-a}{\sqrt{a}}\)

Vậy \(P=\frac{1-a}{\sqrt{a}}\)với a > 0 và \(a\ne1\)

b) Do a > 0 và a khác 1 nên P < 0 khi và chỉ khi :

\(\frac{1-a}{\sqrt{a}}< 0\Leftrightarrow1-a< 0\Leftrightarrow a>1\)

Ta có: \(x^2+\frac{1}{x^2}=14\)(1)

=> \(x^2+\frac{1}{x^2}+2=16\)

<=> \(\left(x+\frac{1}{x}\right)^2=16\)

<=> \(x+\frac{1}{x}=4\) (Vì x > 0)

<=> \(\left(x+\frac{1}{x}\right)^3=4^3\)

<=> \(x^3+3x+\frac{3}{x}+\frac{1}{x^3}=64\)

<=> \(x^3+\frac{1}{x^3}=64-3\left(x+\frac{1}{x}\right)\)

<=> \(x^3+\frac{1}{x^3}=64-3.4=52\) (2)

Từ (1) và (2) nhân vế theo vế:

\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=14.52=728\)

=> \(x^5+\frac{1}{x}+x+\frac{1}{x^5}=728\)

=> \(x^5+\frac{1}{x^5}=728-4=724\)

Để đồ thị hàm số y = (2m - 5)x - 3m + 4 đi qua P thì :

 \(-3=\left(2m-5\right).2-3m+4\)

\(\Leftrightarrow4m-10-3m+7=0\)

\(\Leftrightarrow m=3\)

Vậy ...