X=2 ,Y=3 nha!

x=2,y=3

\(\left|x+\frac{1}{x}\right|=3x-1\)

\(\orbr{\begin{cases}x+\frac{1}{x}=3x-1\\-x-\frac{1}{x}=3x-1\end{cases}}\)

\(\orbr{\begin{cases}x+\frac{1}{x}-3x+1=0\\-x-\frac{1}{x}-3x+1=0\end{cases}}\)

\(\orbr{\begin{cases}-2x+\frac{1}{x}+1=0\\-4x-\frac{1}{x}+1=0\end{cases}}\)

\(\orbr{\begin{cases}-2x^2+1+x=0\\-4x^2-1+x=0\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{1}{2};x=1\\x=\frac{1-\sqrt{15t}}{8}\end{cases}}\)

| x + \(\frac{1}{3}\)| = 3x - 1

\(\Rightarrow\)x + \(\frac{1}{3}\)= \(\pm\)( 3x - 1 )

TH1 : x + \(\frac{1}{3}\)= 3x - 1

\(\Rightarrow\)2x = \(\frac{4}{3}\)

\(\Rightarrow\)x = \(\frac{2}{3}\)

TH2 : x + \(\frac{1}{3}\)= - 3x + 1

\(\Rightarrow\)4x = \(\frac{2}{3}\)

\(\Rightarrow\)x = \(\frac{1}{6}\)

a.

Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow1+2sinx.cosx=t^2\Rightarrow2sinx.cosx=t^2-1\)

Phương trình trở thành:

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

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Rightarrow\left[{}\begin{matrix}t=2>\sqrt{2}\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(1+\dfrac{sinx}{cosx}=2\sqrt{2}sinx\)

\(\Rightarrow sinx+cosx=2\sqrt{2}sinx.cosx\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin2x\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin2x\)

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

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

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)

c.

\(\Leftrightarrow1+sinx+cosx+sinx.cosx=2\)

\(\Leftrightarrow sinx+cosx+sinx.cosx=1\)

Đặt \(sinx+cosx=t\in\left[-\sqrt[]{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(t+\dfrac{t^2-1}{2}=1\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=1\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

Anh nghĩ với bài kiểm tra em nên tự làm nhé. 

a.

\(90^0< a< 180^0\Rightarrow cosa< 0\)

\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{2\sqrt{2}}{3}\)

\(tana=\dfrac{sina}{cosa}=-\dfrac{\sqrt{2}}{4}\)

b.

\(0< a< 90^0\Rightarrow cosa>0\)

\(\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{4}{5}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{3}{4}\)

\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)

c.

\(A=\dfrac{\dfrac{sina}{cosa}+\dfrac{3cosa}{sina}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\dfrac{sin^2a+3cos^2a}{sin^2a+cos^2a}=1+2cos^2a=\dfrac{17}{8}\)

d.

\(A=\dfrac{\dfrac{cosa}{sina}+\dfrac{3sina}{cosa}}{\dfrac{2cosa}{sina}+\dfrac{sina}{cosa}}=\dfrac{cos^2a+3sin^2a}{2cos^2a+sin^2a}=\dfrac{cos^2a+3\left(1-cos^2a\right)}{2cos^2a+\left(1-cos^2a\right)}\)

\(=\dfrac{3-2cos^2a}{1+cos^2a}=\dfrac{19}{13}\)

e.

\(B=\dfrac{\dfrac{3cosa}{sina}+\dfrac{2sina}{cosa}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\dfrac{3cos^2a+2sin^2a}{sin^2a+cos^2a}=3\left(1-sin^2a\right)+2sin^2a\)

\(=3-sin^2a=\dfrac{26}{9}\)

f.

\(C=\dfrac{\dfrac{2sina}{cosa}+\dfrac{3cosa}{cosa}}{\dfrac{sina}{cosa}+\dfrac{cosa}{cosa}}=\dfrac{2tana+3}{tana+1}=\dfrac{7}{3}\)

g.

\(C=\dfrac{\dfrac{3sina}{sina}-\dfrac{4cosa}{sina}}{\dfrac{cosa}{sina}-\dfrac{2sina}{sina}}=\dfrac{3-4cota}{cota-2}=1+\sqrt{5}\)

Gọi \(M\left(x;y\right)\) là 1 điểm bất kì trên (E) \(\Rightarrow\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\) (1)

Gọi \(M'\left(x';y'\right)\) là ảnh của M qua phép tịnh tiến \(\overrightarrow{v}\Rightarrow M'\in\left(E'\right)\) với (E') là ảnh của (E) qua phép tịnh tiến nói trên

\(\left\{{}\begin{matrix}x'=x+3\\y'=y-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=x'-3\\y=y'+2\end{matrix}\right.\)

Thế vào (1):

\(\dfrac{\left(x'-3\right)^2}{16}+\dfrac{\left(y'+2\right)^2}{9}=1\)

Hay pt (E') có dạng: \(\dfrac{\left(x-3\right)^2}{16}+\dfrac{\left(y+2\right)^2}{9}=1\)

Tả cảnh vật thiên nhiên hay tả cảnh gì vậy bạn?

Tả thiên nhiên. Nhanh lên!

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