\(a,tan\alpha=\dfrac{1}{3}\Rightarrow\dfrac{sin\alpha}{cos\alpha}=\dfrac{1}{3}\\ \Rightarrow cos\alpha=3sin\alpha\\Mà:sin^2\alpha+cos^2\alpha=1\\ \Leftrightarrow sin^2\alpha+\left(3sin\alpha\right)^2=1\\ \Leftrightarrow10sin^2\alpha=1\\ \Leftrightarrow sin^2\alpha=\dfrac{1}{10}\\ \Rightarrow sin\alpha=\sqrt{\dfrac{1}{10}}\approx0,3162\\ cos\alpha=\sqrt{1-\dfrac{1}{10}}=\sqrt{\dfrac{9}{10}}\approx0,9487\)

\(b,cot\alpha=\dfrac{3}{4}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\dfrac{3}{4}\Leftrightarrow cos\alpha=\dfrac{3}{4}sin\alpha\\ Ta.có:cos^2\alpha+sin^2\alpha=1\\ \Leftrightarrow\left(\dfrac{3}{4}sin\alpha\right)^2+sin^2\alpha=1\\ \Leftrightarrow\dfrac{9}{16}sin^2\alpha+sin^2\alpha=1\\ \Leftrightarrow\dfrac{25}{16}sin^2\alpha=1\\ \Leftrightarrow sin^2\alpha=1:\dfrac{25}{16}=\dfrac{16}{25}\\ \Rightarrow sin\alpha=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\\ Vậy:cos\alpha=\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)

a) 

\(\tan\alpha=\dfrac{1}{3}\Leftrightarrow\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{1}{3}\Rightarrow3\sin\alpha=\cos\alpha\)

Ta có:

 \(\sin^2\alpha+\cos^2\alpha=1\\ \Leftrightarrow\sin^2\alpha+\left(3\sin\alpha\right)^2=1\\ \Leftrightarrow10\sin^2\alpha=1\\ \Leftrightarrow\sin^2\alpha=\dfrac{1}{10}\)

Vì \(\alpha\) là một góc nhọn => \(\alpha>0.\Rightarrow\sin\alpha=\dfrac{\sqrt{10}}{10}=0,3162\)

\(\cos\alpha=3\sin\alpha=\dfrac{3\sqrt{10}}{10}\approx0,9487\)

b) \(\cot\alpha=\dfrac{3}{4}\Leftrightarrow\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{3}{4}\Rightarrow4\cos\alpha=3\sin\alpha\Leftrightarrow\cos\alpha=\dfrac{3}{4}\sin\alpha\)

Ta có:

\(\sin^2\alpha+\cos^2\alpha=1\\ \Leftrightarrow\sin^2\alpha+\left(\dfrac{3}{4}\sin\alpha\right)^2=1\\ \Leftrightarrow\dfrac{25}{16}\sin^2\alpha=1\\ \Leftrightarrow\sin^2\alpha=\dfrac{16}{25}\)

Vì \(\alpha\) là một góc nhọn => \(\alpha>0\) => \(\sin\alpha=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}\approx0,8\)

\(\cos\alpha=\dfrac{3}{4}\sin\alpha=\dfrac{3}{4}.0,8=\dfrac{3}{5}=0,6\)

ĐKXĐ: \(\dfrac{x-1}{x+1}>=0\)

=>\(\left[{}\begin{matrix}x>=1\\x< -1\end{matrix}\right.\)

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

=>\(\dfrac{x-1}{x+1}=4\)

=>4x+4=x-1

=>3x=-5

=>x=-5/3(nhận)

\(5x^2+4x-1=0\)

=>\(x_1\cdot x_2=-\dfrac{1}{5};x_1+x_2=-\dfrac{4}{5}\)

\(A=\left(5x_1-3x_2\right)\left(5x_2-3x_1\right)\)

\(=25x_1x_2-15x_1^2-15x_2^2+9x_1x_2\)

\(=34x_1x_2-15\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\)

\(=34\cdot\dfrac{-1}{5}-15\left[\left(-\dfrac{4}{5}\right)^2-2\cdot\dfrac{-1}{5}\right]\)

\(=-\dfrac{34}{5}-15\cdot\left[\dfrac{16}{25}+\dfrac{2}{5}\right]\)

\(=\dfrac{-34}{5}-15\cdot\dfrac{26}{25}\)

\(=\dfrac{-34}{5}-\dfrac{78}{5}=-\dfrac{112}{5}\)

\(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=\left(-\dfrac{4}{5}\right)^2-2\cdot\dfrac{-1}{5}\)

\(=\dfrac{16}{25}+\dfrac{2}{5}=\dfrac{16+10}{25}=\dfrac{26}{25}\)

\(C=\left|x_1-x_2\right|\)

\(=\sqrt{\left(x_1-x_2\right)^2}\)

\(=\sqrt{x_1^2+x_2^2-2x_1x_2}=\sqrt{\dfrac{26}{25}-2\cdot\dfrac{-1}{5}}=\dfrac{6}{5}\)

\(D=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)

\(=\left(-\dfrac{4}{5}\right)^3-3\cdot\dfrac{-1}{5}\cdot\dfrac{-4}{5}\)

\(=-\dfrac{64}{125}-3\cdot\dfrac{4}{25}\)

\(=\dfrac{-64}{125}-\dfrac{12}{25}=\dfrac{-124}{125}\)

\(\sqrt{-\dfrac{5}{2x+4}}\) có nghĩa khi và chỉ khi:\(\left\{{}\begin{matrix}2x+4\ne0\\-\dfrac{5}{2x+4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\2x+4< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\2x< -4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x< -2\end{matrix}\right.\Leftrightarrow x< -2\)

\(\left(\sqrt[3]{\dfrac{1}{9}}+4\cdot\sqrt[3]{\dfrac{1}{72}}-\sqrt[3]{4}\right)\left(\sqrt[3]{72}+\sqrt[3]{96}+\sqrt[3]{128}\right)\)

\(=\left(\dfrac{1}{3}\cdot\sqrt[3]{3}+4\cdot\dfrac{1}{6}\cdot\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(2\sqrt[3]{9}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=\left(\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(6\sqrt[3]{3}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=6\cdot3+2\sqrt[3]{36}+4\sqrt[3]{6}-12\sqrt[3]{\dfrac{3}{2}}-4\sqrt[3]{6}-8\)

\(=10+12\sqrt[3]{\dfrac{1}{6}}-6\sqrt[3]{12}\)

Ta có: \(\sqrt{x}+3\ge3\forall x\ge0\)

Nên: \(A=\dfrac{-3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\) 

Dấu "=" xảy ra:

\(\dfrac{-3}{\sqrt{x}+3}=-1\)

\(\Leftrightarrow x=0\)

Vậy: \(A_{min}=-1.khi.x=0\)

\(\sqrt{3x+1}+2\sqrt{x+3}=3\sqrt{5x-1}\)

=>\(\sqrt{3x+1}-2+2\sqrt{x+3}-4=3\sqrt{5x-1}-6\)

=>\(\dfrac{3x+1-4}{\sqrt{3x+1}+2}+2\left(\sqrt{x+3}-2\right)-3\left(\sqrt{5x-1}-2\right)=0\)

=>\(\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}+2\cdot\dfrac{x+3-4}{\sqrt{x+3}+2}-3\cdot\dfrac{5x-1-4}{\sqrt{5x-1}+2}=0\)

=>\(\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{2}{\sqrt{x+3}+2}-\dfrac{15}{\sqrt{5x-1}+2}\right)=0\)

=>x-1=0

=>x=1