Bài 2: 

Thay x=3 và y=-5 vào (d), ta được:

b-6=-5

hay b=1

a: tan B=2

=>\(\frac{\sin B}{cosB}=2\)

=>sin B=2*cosB

\(A=\frac{\sin B+cosB}{\sin B-cosB}\)

\(=\frac{2\cdot cosB+cosB}{2\cdot cosB-cosB}=\frac{3\cdot cosB}{cosB}=3\)

b: \(C=\sin^2a+2\cdot\sin a\cdot cosa-3\cdot cos^2a\)

\(=\left(\sin^2a+2\cdot\sin a\cdot cosa+cos^2a\right)-4\cdot cos^2a\)

\(=\left(\sin a+cosa\right)^2-4\cdot cos^2a=\left(3\cdot cosa\right)^2-4\cdot cos^2a=5\cdot cos^2a\)

c: \(D=\frac{\sin^2a-\sin a\cdot cosa-cos^2a}{2\cdot\sin a\cdot cosa}\)

\(=\frac{\left(2\cdot cosa\right)^2-2\cdot cosa\cdot cosa-cos^2a}{2\cdot2\cdot cosa\cdot cosa}\)

\(=\frac{4\cdot cos^2a-3\cdot cos^2a}{4\cdot cos^2a}=\frac{4-3}{4}=\frac14\)

Bài 5: 

a: Xét ΔBEC và ΔADC có 

\(\widehat{C}\) chung

\(\widehat{EBC}=\widehat{DAC}\)

Do đó: ΔBEC\(\sim\)ΔADC

có☹

\(\sqrt{12}=\sqrt{x^2+12x+13}\)

\(\Leftrightarrow12=x^2+12x+13\)

\(\Leftrightarrow x^2+12x+36-35=0\)

\(\Leftrightarrow\left(x+6\right)^2-35=0\)

\(\Leftrightarrow\left(x+6+\sqrt{35}\right)\left(x+6-\sqrt{35}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-6-\sqrt{35}\\x=\sqrt{35}-6\end{cases}}\)

Đặt \(\hept{\begin{cases}\sqrt[3]{x-16}=a\\\sqrt[3]{x+13}=b\end{cases}}\)

\(\Rightarrow b^3-a^3=29\)

Từ đó ta có hệ \(\hept{\begin{cases}1+a=b\\b^3-a^3=29\end{cases}}\)

Thế pt đầu vào pt sau ta được

\(a^3+3a^2+3a+1-a^3=29\)

\(\Leftrightarrow3a^2+3a-28=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{-3+\sqrt{345}}{6}\\a=\frac{-3-\sqrt{345}}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}b=\frac{3+\sqrt{345}}{6}\\b=\frac{3-\sqrt{345}}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{31\sqrt{345}+27}{18}\\x=\frac{-31\sqrt{345}+27}{18}\end{cases}}\)

\(12-2\sqrt{35}\)

\(=\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{35}\)

\(=\left(\sqrt{5}+\sqrt{7}\right)^2\)

\(7+\sqrt{40}\)

\(=\left(\sqrt{5}\right)^2+\left(\sqrt{2}\right)^2+2\sqrt{10}\)

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

a) \(12-2\sqrt{35}=\left(\sqrt{5}\right)^2-2\sqrt{5.7}+\left(\sqrt{7}\right)^2=\left(\sqrt{5}-\sqrt{7}\right)^2\)

b) \(7+\sqrt{40}=7+\sqrt{4.10}=7+2\sqrt{10}=\left(\sqrt{5}\right)^2+2\sqrt{5.2}+\left(\sqrt{2}\right)^2=\left(\sqrt{5}+\sqrt{2}\right)^2\)