\(x-7=\left(\sqrt{x}\right)^2-\left(\sqrt{7}\right)^2=\left(\sqrt{x}-\sqrt{7}\right)\left(\sqrt{x}+\sqrt{7}\right)\)( \(x\ge0\))

\(x-6\sqrt{x}+9=\left(\sqrt{x}\right)^2-2.3.\sqrt{x}+3^2=\left(\sqrt{x}-3\right)^2\)( \(x\ge0\))

Em mới lớp 8 nên không dám chắc ạ :(

\(4x^2+4xy-y^2=0\)

Rút gọn thừa số chung : \(-y^2+4xy+4x^2=0\)

Dơn giản biểu thức : \(-y^2+4xy+4x^2-0=0\)

Giải phương trình : \(-\left(y^2-4xy-4x^2\right)2=0\)

Giải phương trình : \(y^2-4xy-4x^2=0\)

Đặt:    \(B=\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}\)

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

=>   \(B^2=14+2\sqrt{49-5}\)

=>   \(B^2=14+2\sqrt{44}\)

=>   \(A=\frac{\sqrt{14+4\sqrt{11}}}{7+2\sqrt{11}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

=>   \(A=\sqrt{\frac{2}{7+2\sqrt{11}}}-\left(\sqrt{2}-1\right)\)

=>   \(A=\sqrt{\frac{2}{7+2\sqrt{11}}}-\sqrt{2}+1\)

ĐỀ BÀI CHẮC SAI RỒI PHẢI DƯỚI MẪU PHẢI LÀ    \(\sqrt{7+2\sqrt{11}}\)    THÌ LÚC ĐÓ BIỂU THỨC A RA ĐẸP HƠN !!!!

NẾU SỬA ĐỀ BÀI NHƯ TRÊN:

=>    \(A=\frac{\sqrt{2}.\sqrt{7+2\sqrt{11}}}{\sqrt{7+2\sqrt{11}}}-\left(\sqrt{2}-1\right)\)

=>   \(A=\sqrt{2}-\sqrt{2}+1\)

=>   \(A=1\)

ĐÓ BÂY GIỜ RA A  = 1 RẤT ĐẸP

<=>   \(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)

<=>   \(x^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{12-6\sqrt{2+\sqrt{3}}+6\sqrt{2+\sqrt{3}}-3\left(2+\sqrt{3}\right)}\)

<=>   \(x^2=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-2\sqrt{12-6-3\sqrt{3}}\)

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

<=>   \(x^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{2}.\sqrt{12-6\sqrt{3}}\)

<=>   \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}.\sqrt{\left(3-\sqrt{3}\right)^2}\)

<=>   \(x^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left(3-\sqrt{3}\right)\)

<=>   \(x^2=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)

<=>   \(x^2=8-4\sqrt{2}\)

<=>   \(8-x^2=4\sqrt{2}\)

<=>   \(\left(8-x^2\right)^2=\left(4\sqrt{2}\right)^2\)

<=>   \(x^4-16x^2+64=32\)

<=>   \(x^4-16x^2=-32\)

VẬY    \(x^4-16x^2=-32\)

*** ĐÂY LÀ 1 BÀI TOÁN RẤT CỔ RỒI !!!!!!

+) \(x+y+xy=8\Leftrightarrow\left(x+1\right)\left(y+1\right)=9\)

+) Đặt: \(a=\sqrt{x+1};b=\sqrt{y+1}\)

+) \(P=\frac{\sqrt{x+1}+\sqrt{y+1}}{\left(x+1\right)\left(y+1\right)-\left(x+1\right)-\left(y+1\right)+2}=\frac{a+b}{11-a^2-b^2}\)

\(\ge\frac{2\sqrt{ab}}{11-2ab}=\frac{2\sqrt{3}}{11-2\cdot3}=\frac{2\sqrt{3}}{5}\)

Dấu = xảy ra khi x = y = 2

+) \(P^2=\frac{x+y+8}{\left(xy+1\right)^2}=\frac{16-xy}{\left(xy+1\right)^2}\le\frac{16}{1}=4\)

\(\Rightarrow P\le4\)

Dấu = xảy ra khi \(\orbr{\begin{cases}x=8;y=0\\x=0;y=8\end{cases}}\)

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

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

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

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

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

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

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

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

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

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

\(=1-3=-2\)