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

\(=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}.\left(\frac{\left(3\sqrt{2}-2\sqrt{3}\right)^2}{\sqrt{6}}\right)\)

\(=\frac{\sqrt{3}+\sqrt{2}}{-1}.\left(\frac{30-12\sqrt{6}}{\sqrt{6}}\right)\)

\(=\frac{\sqrt{6}\left(\sqrt{150}-12\right)\left(\sqrt{3}+\sqrt{2}\right)}{-\sqrt{6}}\)

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

\(=-\left(5\sqrt{18}+5\sqrt{12}-12\sqrt{3}-12\sqrt{2}\right)\)

\(=-\left(15\sqrt{2}+10\sqrt{3}-12\sqrt{3}-12\sqrt{2}\right)\)

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

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

VẬY   \(VT=2\sqrt{3}-3\sqrt{2}\)

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

\(=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}.\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)

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

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

\(=-\left(3-2\right)=-1\)

\(x^2-2\sqrt{2}-6=0\)

\(\Leftrightarrow\left(x+\sqrt{2}\right)\left(x-3\sqrt{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\sqrt{2}\\x=3\sqrt{2}\end{cases}}\)

KL:...

\(x^2-2\sqrt{2x}-6=0\)

\(\left(x+\sqrt{2}\right)\left(x-3\sqrt{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\sqrt{2}=0\\x-3\sqrt{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\sqrt{2}\\x=3\sqrt{2}\end{cases}}}\)

\(\sin\alpha=\frac{2}{5}\)

\(\Rightarrow\cos\alpha=\sqrt{1-\sin^2\alpha}\)

\(=\sqrt{1-\frac{4}{25}}\)

\(=\sqrt{\frac{21}{25}}=\)\(\frac{\sqrt{21}}{5}\)

\(\Rightarrow\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{5}:\frac{\sqrt{21}}{5}=\frac{2}{\sqrt{21}}\)và \(\cot\alpha=\frac{\sqrt{21}}{2}\)

2. Tương tự a)

\(\cos B=\sqrt{1-\sin^2B}\)

\(=\sqrt{1-\frac{1}{4}}\)

\(=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)

\(\tan B,\cot B\)bạn tự tính nốt.

\(sin\alpha=0,4\Rightarrow sin^2\alpha=0,16\Rightarrow cos^2\alpha=1-sin^2\alpha=1-0,16=0,84\Rightarrow cos\alpha=\frac{\sqrt{21}}{5}\)

\(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{0,4}{\frac{\sqrt{21}}{5}}=\frac{2\sqrt{21}}{21}\)

\(cot\alpha=1:sin\alpha=1:\frac{2\sqrt{21}}{21}=\frac{21}{2\sqrt{21}}\)

A: 3.65028154 / B: 3.445719569 / C: 0.4743791865

\(\sqrt{7+\sqrt{40}}\)

\(=\sqrt{7+2\sqrt{2}.\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{5}\)

\(\)

a) vẽ phân giác BD của góc ABC. theo tính chất đường phân giác ta có \(\frac{BA}{BC}=\frac{DA}{DC}\Rightarrow\frac{DA}{BA}=\frac{DC}{BC}\left(1\right)\)

theo tính chất dãy tỉ số bằng nhau ta có \(\frac{DA}{BA}=\frac{DC}{BC}=\frac{DA+DC}{BA+BC}=\frac{AC}{AB+BC}\left(2\right)\)

tam giác BAD vuông tại A nên \(\tan\widehat{ABD}=\frac{DA}{BA}\left(3\right)\)

từ (2) và (3) ta có \(\tan\widehat{ABD}=\frac{AC}{AB+BC}\)hay \(\tan\frac{\widehat{B}}{2}=\frac{AC}{AB+BC}\)

b) áp dụng kết quả phần (a) ở trên, giả sử tam giác ABC vuông cân tại A, AB=a khi đó

\(\tan\frac{\widehat{B}}{2}=\tan22^030'=\frac{AC}{AB+BC}=\frac{a}{a+a\sqrt{2}}=\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\)

áp dụng kết quả ở phần (a) ở trên, giả sử tam giác ABC vuông tại A, \(\widehat{B}=30^o;AC=a\)khi đó

\(\tan\frac{\widehat{B}}{2}=\tan15^o=\frac{AC}{AB+BC}=\frac{a}{a\sqrt{3}+2a}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)

mọi người ơi giúp em câu này với ạ

a) \(ĐKXĐ:x>0;x\ne4\)

Ta có : \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)

\(=\left[\frac{\sqrt{x}.\sqrt{x}-4x}{\sqrt{x}.\left(\sqrt{x}-2\right)}\right]\cdot\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

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

b) Ta có : \(x-1=10-4\sqrt{6}=\left(\sqrt{6}-2\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}-2\right)^2+1}\)

......

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

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

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

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

\(=\left(3+\sqrt{5}\right).\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\)

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

\(=\frac{\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2}{2}=\frac{\left[\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\right]^2}{2}\)

\(=\frac{\left(5-1\right)^2}{2}=\frac{4^2}{2}=8\)