a, \(\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)

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

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

\(=4-3=1\)

b, \(\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}\)(bạn xem lại đề có bị sai hay không)

a, \(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\) (ĐKXĐ: \(x\ne1,x\ge0\))

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

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

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

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

b, \(A-\frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}\)\(=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=-\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=-\frac{\left(\sqrt{x}+1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)

\(\Rightarrow A-\frac{1}{3}< 0\Leftrightarrow A< \frac{1}{3}\)

c, ĐKXĐ: \(x\ge0,x\ne1\)

Ta có: x = \(19-8\sqrt{3}\)(TMĐK) \(\Leftrightarrow\sqrt{x}=\sqrt{19-8\sqrt{3}}\Leftrightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{3}\right)^2}\Leftrightarrow\sqrt{x}=4-\sqrt{3}\)

Thay \(\sqrt{x}=4-\sqrt{3}\)vào A ta có:

\(A=\frac{4-\sqrt{3}}{\left(4-\sqrt{3}\right)^2+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{19-8\sqrt{3}+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)

Vậy với \(x=19-8\sqrt{3}\)thì \(A=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)

\(A=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{16.5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

\(B=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)

\(=2\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}+3\sqrt{22}=22\)

ko bit

\(5\sqrt{a}-4b\sqrt{25^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-4b.25a\sqrt{a}+5a.4b\sqrt{a}-6\sqrt{a}\)

\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\)

\(=-\sqrt{a}\)

Mn trả lời nhanh nhanh giùm em với ạ. Em đang cần gấp...

- Ta có: \(\sin\alpha+\cos\alpha=\frac{7}{5}\)  

        \(\Rightarrow\sin\alpha=\frac{7}{5}-\cos\alpha\)

- Theo tỉ số lượng giác của óc nhọn, ta có:

         \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\left(\frac{7}{5}-\cos\alpha\right)^2+\cos^2\alpha=1\)

\(\Leftrightarrow\frac{49}{25}-\frac{14}{5}\cos\alpha+\cos^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow50\cos^2\alpha-70\cos\alpha+48=0\)

\(\Leftrightarrow25\cos^2\alpha-35\cos\alpha+24=0\)

\(\Leftrightarrow\left(5\cos\alpha-4\right)\left(5\cos\alpha-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5\cos\alpha-4=0\\5\cos\alpha-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\cos\alpha=\frac{4}{5}\\\cos\alpha=\frac{3}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\sin\alpha=\frac{7}{5}-\cos\alpha=\frac{7}{5}-\frac{4}{5}=\frac{3}{5}\\\sin\alpha=\frac{7}{5}-\cos\alpha=\frac{7}{5}-\frac{3}{5}=\frac{4}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}\\\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\end{cases}}\)

Kết luận: Vậy..........

\(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\sin^2\alpha+\left(\frac{7}{5}-\sin\alpha\right)^2=1\)

\(\Rightarrow25\sin^2\alpha-35\sin\alpha+12=0\)

\(\Rightarrow\left(5\sin\alpha-4\right)\left(5\sin\alpha-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sin\alpha=\frac{4}{5}\\\sin\alpha=\frac{3}{5}\end{cases}}\)

Nếu \(\sin\alpha=\frac{4}{5}\)thì \(\cos\alpha=\frac{3}{5}\Rightarrow\tan\alpha=\frac{4}{3}\)

Nếu \(\sin\alpha=\frac{3}{5}\)thì \(\cos\alpha=\frac{4}{5}\Rightarrow\tan\alpha=\frac{3}{4}\)

Tk cho mk bạn nhá