Gọi chiều rộng là x

Chiều dài là x+15

Theo đề, ta có phương trình:

\(\left(x+5\right)\left(x+12\right)=x\left(x+15\right)+80\)

\(\Leftrightarrow x^2+17x+60-x^2-15x=80\)

=>2x+60=80

=>x=10

Vậy: Chiều rộng là 10m

Chiều dài là 25m

Gọi độ dài quãng đường là x

Theo đề, ta có: 

\(\dfrac{x}{42}-\dfrac{x}{46}=\dfrac{3}{4}\)

hay x=362,25(km)

\(x=\left(\dfrac{1}{2}\right)^3:\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{3-1}=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)

1/4

\(\dfrac{1}{4}\)

Ta có: \(\frac{x+2}{y+10}\)\(=\)\(\frac{1}{5}\)\(\Rightarrow\)\(5\left(x+2\right)=y+10\)(1)

             \(y-3x=2\)\(\Rightarrow\)\(y+2=3x\)                              (2)

Thay (2) vào (1) ta có:

\(5\left(x+2\right)=\left(y+2\right)+8\)

\(5x+10=3x+8\)

\(5x-3x=8-10\)

\(2x=-2\)

\(x=-2:2\)

\(x=-1\)

Vậy: x=-1

Chúc bạn làm bài tốt!

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

c.

\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)

\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

2.

\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)

\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)

\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)

Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)

\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)

Help

2:

1+cot^2a=1/sin^2a

=>1/sin^2a=1681/81

=>sin^2a=81/1681

=>sin a=9/41

=>cosa=40/41

tan a=1:40/9=9/40

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