Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
- \(=\frac{\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right)}{\sqrt{35}}=\sqrt{5}+\sqrt{7}\)
- \(=\frac{4\sqrt{2}-3\sqrt{3}+1}{\sqrt{3}\sqrt{2}}=\frac{4}{\sqrt{3}}+\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{6}}\)
- \(=\frac{\left(3\sqrt{11}-3\sqrt{3}-\sqrt{11}\right)}{\sqrt{11}}+3\sqrt{2}=\frac{\left(2\sqrt{11}-3\sqrt{3}\right)}{\sqrt{11}}+3\sqrt{2}\)\(=\frac{2\sqrt{11}-3\sqrt{3}+3\sqrt{22}}{\sqrt{11}}\)
câu c bạn làm nhầm đề bài r kìa Hoàng Anh Tuấn
\(\sqrt{18}=3\sqrt{2}\) chứ sao lại bằng \(3\sqrt{3}\)đc
- \(\left(\sqrt{5^2.7}+\sqrt{7^2.5}\right):\sqrt{35}=\sqrt{35}\left(\sqrt{5}+\sqrt{7}\right):\sqrt{35}=\sqrt{5}+\sqrt{7}\)
- \(\left(\sqrt{2^2.8}-3\sqrt{3}+1\right):\sqrt{2.3}=\frac{4}{\sqrt{3}}-\frac{3}{\sqrt{2}}+\frac{1}{\sqrt{6}}\)
- \(\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right):\sqrt{11}+3\sqrt{2}\sqrt{11}\)\(=\frac{\left(2\sqrt{11}-3\sqrt{2}\right)}{\sqrt{11}}+3\sqrt{2}\sqrt{11}\)\(=\frac{2\sqrt{11}-3\sqrt{2}+33\sqrt{2}}{\sqrt{11}}=\frac{2\sqrt{11}-30\sqrt{2}}{\sqrt{11}}\)
Hoàng Anh Tuấn : mik vẫn chưa hiểu câu 2 , 3 b ra thế nào ? xin b hãy giải theo 1 cách dễ hiểu hay giảng cho mik đc ko ạ !!!
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a) Ta có: \(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}+3\right)\left(\sqrt{2}-\sqrt{3}-3\right)}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-4-2\sqrt{6}}\)
\(=\frac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{-2\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}+\sqrt{3}\right)}{-\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{3\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}-3\right)}{2}\)
b) Ta có: \(\left(\frac{4}{\sqrt{5}+1}-\frac{4}{\sqrt{5}-1}\right):\sqrt{3+2\sqrt{2}}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\frac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\right):\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)
\(=\left(\frac{4\left(\sqrt{5}-1\right)}{4}-\frac{4\left(\sqrt{5}+1\right)}{4}\right):\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\sqrt{5}-1-\sqrt{5}-1\right):\left|\sqrt{2}+1\right|\)
\(=-\frac{2}{\sqrt{2}+1}\)(Vì \(\sqrt{2}+1>0\))
\(=-\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=-2\left(\sqrt{2}-1\right)\)
\(=-2\sqrt{2}+2\)
\(3\sqrt{2a}-\sqrt{2.3^2a.a^2}-\frac{1}{4}\sqrt{8^2.2a}=3\sqrt{2a}-3a\sqrt{2a}-2\sqrt{2a}=\sqrt{2a}-3a\sqrt{2a}\)
\(\left(1-3a\right)\sqrt{2a}\)
nếu là phương trình :
\(\sqrt{2a}\left(1-3a\right)=0\Leftrightarrow\left(1-3a\right)=0\Leftrightarrow1-3a=0\Leftrightarrow a=\frac{1}{3}\)
Thế cái đề này là nó yêu cầu làm gì?