Cho t làm bừa phát :v

\(B=\frac{15}{\sqrt{6}+1}-\frac{6}{\sqrt{6}-2}\)

\(B=\frac{15\sqrt{6}-15}{\sqrt{6^2}-1}-\frac{6\sqrt{6}+12}{\sqrt{6^2}-2^2}\)

\(B=\frac{15\left(\sqrt{6}-1\right)}{\sqrt{6^2}-1}-\frac{6\left(\sqrt{6}+2\right)}{\sqrt{6^2}-2^2}\)

\(B=\frac{15\left(\sqrt{6}-1\right)}{6-1}-\frac{6\left(\sqrt{6}+2\right)}{6-4}\)

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

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

\(B=3\sqrt{6}-3-3\sqrt{6}-6\)

\(B=\left(3\sqrt{6}-3\sqrt{6}\right)+\left(-3-6\right)\)

\(=-9\)

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

bạn làm như này nha:

Từ đpcm  \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)

             \(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

             \(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)      

             \(\Leftrightarrow0=2.\left(\frac{a+b+c}{abc}\right)\)

             \(\Leftrightarrow0=a+b+c\)luôn đúng do giả thuyết cho

                                \(\Rightarrowđpcm\)

A= 5 căn 3 - 3*4 căn 3 +2*5 căn 3 -1/3 *6 căn 3 = căn 3

\(A=5\sqrt{3}-3\sqrt{48}+2\sqrt{75}-\frac{1}{3}\sqrt{108}\)

\(A=5\sqrt{3}-3.\sqrt{4^2.3}+2\sqrt{5^2.3}-\frac{1}{3}.\sqrt{6^2.3}\)

\(A=5\sqrt{3}-12\sqrt{3}+10\sqrt{3}-2\sqrt{3}\)

\(A=\sqrt{3}\)

\(\text{ĐKXĐ: }x\ge0;x\ne\pm1\)

\(2\sqrt{144x+144}-3\sqrt{100x-100}=12\)

\(2\sqrt{144\left(x+1\right)}-3\sqrt{100\left(x-1\right)}=12\)

\(2\sqrt{144}.\sqrt{\left(x+1\right)}-3\sqrt{100}.\sqrt{x-1}=12\)

\(2.12\sqrt{x+1}-3.10\sqrt{x-1}=12\)

\(24\sqrt{x+1}-30\sqrt{x-1}=12\)

\(6.\left(4\sqrt{x+1}-5\sqrt{x-1}\right)=6.2\)

\(4\sqrt{x+1}-5\sqrt{x-1}=2\)

\(\text{Mk bí r}\)

Today, I will lecture about the country of Japan. the reason that I love this country as comics and cuisine. Japanese comics attached to my childhood. I spent a private space to collect them. There are many famous series doraemon cônan. Japanese culinary culture is very rich and rich nutritional value. the Japanese usually use fresh ingredients to make dishes. and in parular they favor fish and vegetables. I would to try the noodles or shushi in Japan. I have read in comics, look at them seems very appealing. Life in Japan is a high-quality life and modern. The Japanese are very careful and always demanding in all things. they are very hospitable and always help others, especially for international students.  other reason I love for a landscape that is. for example Mount Fuji or cherry. the famous temples in kyoto

E có thể tham khảo tại link :

Introduction to japan

https://123doc.org/document/3836326-introduction-to-japan.htm

_Minh ngụy_

1. 

a. Xưng khiêm là khi dùng đại từ để nói về chính bản thân mình thì khiêm tốn. 

Hô tôn: nói, gọi người khác với thái độ tôn trọng, đặt họ ở vị trí trên.

b. Phương châm hội thoại được sử dụng trong câu trên là phương châm lịch sự.

c. Vận dụng câu thành ngữ với nhiều đối tượng khác nhau như: người mới quen, người trên, bạn bè...

\(22-2\sqrt{x+5}+1=-x^2-9x\)

\(23-2\sqrt{x+5}=-x^2-9x\)

\(2\sqrt{x+5}=23+x^2+9x\)

\(4\left(x+5\right)=\left(23+x^2+9x\right)^2\)

\(4x+20=529+x^4+81x^2+46x^2+18x^3+261x\)

\(x^4+18x^3+81x^2+257x+509=0\)

bấm máy thì ra nha

\(b,P=\sqrt{12-4\sqrt{3}}\)

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

\(c.\sqrt{\left(2\sqrt{3}\right)^2+4\sqrt{3}+1}-\sqrt{\left(2\sqrt{3}\right)^2-4\sqrt{3}+1}\)

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

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

\(=2\)