\(k\in Z\)

a.

\(cos\left(x-2\right)=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=arccos\left(\dfrac{2}{5}\right)+k2\pi\\x-2=-arccos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+arccos\left(\dfrac{2}{5}\right)+k2\pi\\x=2-arcos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3>1\left(vô-nghiệm\right)\end{matrix}\right.\)

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

f.

\(\Leftrightarrow cosx=-\dfrac{1}{2}\)

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

h.

\(cos\left(3x+10^0\right)=-1\)

\(\Leftrightarrow3x+10^0=180^0+k360^0\)

\(\Leftrightarrow3x=170^0+k360^0\)

\(\Leftrightarrow x=\dfrac{1}{3}.170^0+k120^0\)

j.

\(cos\left[cos\left(x+2\right)\right]=1\)

\(\Leftrightarrow cos\left(x+2\right)=k2\pi\)

Do \(-1\le cos\left(x+2\right)\le1\Rightarrow-1\le k2\pi\le1\)

\(\Rightarrow k=0\)

\(\Rightarrow cos\left(x+2\right)=0\)

\(\Rightarrow x+2=\dfrac{\pi}{2}+n\pi\)

\(\Rightarrow x=-2+\dfrac{\pi}{2}+n\pi\)

\(tanx=-tan\dfrac{\pi}{5}\)

\(\Leftrightarrow tanx=tan\left(-\dfrac{\pi}{5}\right)\)

\(\Leftrightarrow x=-\dfrac{\pi}{5}+k\pi\)

Mình quên mất, nó nằm trong khoảng (π/2; π) nha, mình xin lỗi

Gọi số cần tìm là ab, ta có:

ab - ba = 54 ; a = 3b

ab - ba = 54

<=> [10a + b] - [10b + a] = 54

<=> 10a + b - 10b - a = 54

<=> 9a - 9b = 54

<=> 9[a-b] = 54

<=> a - b = 6

Giờ thì khỏe r, ta giải bài tổng tỉ.

a = 3b

=> a - b = 3b - b = 2b = 6

Vậy b = 6/2 = 3

a = 3.3 = 9

Vậy số ab = 93

Đkxđ: \(\hept{\begin{cases}x\ge-\frac{1}{4}\\y\ge2\end{cases}}\)

\(\Leftrightarrow2+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=y\Leftrightarrow2+\frac{1}{2}+\sqrt{x+\frac{1}{2}}=y\Leftrightarrow\sqrt{x+\frac{1}{2}}+\frac{5}{2}=y\)

do x,y nguyên dương nên \(\sqrt{x+\frac{1}{2}}+\frac{5}{2}\)nguyên dương\(\Leftrightarrow\sqrt{x+\frac{1}{2}}=\frac{k}{2}\)(K là số nguyên lẻ, \(k>1\))

\(\Rightarrow x=\frac{k^2-2}{4}\)

do \(k^2\)là số chính phương chia 4 dư 0,1 \(\Rightarrow x=\frac{k^2-2}{4}\notin Z\)

=> ko tồn tại cặp số nguyên dương x,y tmđkđb

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

\(tan\left(\dfrac{3\pi}{2}-\alpha\right)+cot\left(3\pi-\alpha\right)-cos\left(\dfrac{\pi}{2}-\alpha\right)+2.sin\left(\pi+\alpha\right)\)

\(=tan\left(\pi+\dfrac{\pi}{2}-\alpha\right)+cot\left(-\alpha\right)-sin\alpha+2\left(sin\pi.cos\alpha+cos\pi.sin\alpha\right)\)

\(=tan\left(\dfrac{\pi}{2}-\alpha\right)-cot\alpha-sin\alpha+2.-sin\alpha\)

\(=cot\alpha-cot\alpha-3sin\alpha\)

\(=-3sin\alpha\)

\(PT\Leftrightarrow9x^2+16x+96=9x^2+256y^2+576-96xy+768y-144x.\)

\(\Leftrightarrow256y^2-160x-96xy+768y+480=0\)

\(\Leftrightarrow8y^2-5x-3xy+24y+15=0\)

Đến chỗ này phân tích kiểu j được nhỉ

\(x\left(2x-1\right)\left(3x-126\right)=0\Rightarrow\hept{\begin{cases}x=0\\2x-1=0\\3x-126=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\2x=1\\3x=126\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=42\end{cases}}\)