\(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2\sqrt{2}.\sqrt{5}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)

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

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

\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)

\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)

\(A^2=144-80-16\sqrt{5}\)

\(A^2=64-16\sqrt{5}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)

\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)

\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=2+2.\sqrt{2}.\sqrt{10}+10\)

\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)

=> \(A=\sqrt{2}+\sqrt{10}\)

a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

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

\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)

\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)

b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)

\(=-60-144\sqrt{2}+30\sqrt{2}+144\)

\(=84-114\sqrt{2}\)

Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath

Bạn tham khảo link này!

Biến đổi vế trái ta có :

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

= \(\sqrt{2}\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)\)

Đặt A  = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

A^2 = \(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

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

     = \(8+2\sqrt{16-10+2\sqrt{5}}\)

     = \(8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)

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

=> \(\sqrt{2}A=\sqrt{2}\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}=VP\) ( ĐPCM) 

haha        

bn thang tran lm sai bước đưa ra hdt :v đúng là phải 16 - ( 10 + 2can5 )

= 16 - 10 - 2can5

c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)

=> đpcm.

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

áp dụng công thức cho biểu thức A có A>\(2\left(-\sqrt{2}+\sqrt{26}\right)>7\left(1\right)\)

(so sánh bình phương 2 số sẽ ra nha)

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

áp dụng công thức cho biểu thức A ta CM được

A<\(2\left(\sqrt{2}-\sqrt{2-1}+\sqrt{3}-\sqrt{3-1}+...+\sqrt{25}-\sqrt{25-1}\right)\)

=\(2\left(-\sqrt{1}+\sqrt{25}\right)=2\left(-1+5\right)=2\cdot4=8\left(2\right)\)

từ (1) và (2) => ĐPCM

b. tương tự câu a ta CM đc BT đã cho=B>\(2\sqrt{51}-2\)> \(5\sqrt{2}\left(1\right)\)

và B<\(2\sqrt{50}=\sqrt{2}\cdot\sqrt{2\cdot50}=10\sqrt{2}\left(2\right)\)

từ (1) và (2)=>ĐPCM

(bạn nhớ phải biến đổi 1 thành 1/\(\sqrt{1}\) trc khi áp dụng công thức nha)

MỜI BẠN THAM KHẢO

Nhân tử và mẫu của biểu thức với \(\sqrt{m}+\sqrt{n}-\sqrt{m+n}.\)

\(\Rightarrow\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}+\sqrt{m+n}\right)\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+n+2\sqrt{mn}-m-n}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)

Ta có: \(\frac{2\sqrt{mn}}{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}\)

\(=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{\left(\sqrt{m}+\sqrt{n}\right)^2-\left(\sqrt{m+n}\right)^2}=\frac{2\sqrt{mn}.\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{m+2\sqrt{mn}+n-m-n}\)

\(=\frac{2\sqrt{mn}\left(\sqrt{m}+\sqrt{n}-\sqrt{m+n}\right)}{2\sqrt{mn}}=\sqrt{m}+\sqrt{n}-\sqrt{m+n}\)( đpcm )

Áp dụng: Với \(m=2\)và \(n=5\)và \(mn=10\); \(m+n=7\)ta có:

\(\frac{2\sqrt{10}}{\sqrt{2}+\sqrt{5}+\sqrt{7}}=\sqrt{2}+\sqrt{5}-\sqrt{2+5}=\sqrt{2}+\sqrt{5}-\sqrt{7}\)

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+\sqrt{2}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(\sqrt{6}+1\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{\sqrt{2}\left(3\sqrt{6}+3+\sqrt{30}+\sqrt{5}\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+2\sqrt{15}+\sqrt{10}-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+\sqrt{15}+\sqrt{10}-\sqrt{5}}{ }\)

Đề sai rồi bạn