Bạn tự phân tích đa thức thành nhân tử nhé! 

\(1.\)

\(2x^3+x+3=0\)

\(\Leftrightarrow\)  \(\left(x+1\right)\left(2x^2-2x+3\right)=0\)  \(\left(1\right)\)

Vì  \(2x^2-2x+3=2\left(x^2-x+1\right)+1=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>0\)  với mọi  \(x\in R\)

nên từ  \(\left(1\right)\)  \(\Rightarrow\)  \(x+1=0\)  \(\Leftrightarrow\)  \(x=-1\)

1)2x^3+x+3=0=>

Phân tích ra thôi bạn!

\(2\left(x^2+2\right)=5\sqrt{x^3+1}\left(đk:x\ge-1\right)\)

\(\Leftrightarrow2\left[\left(x^2-x+1\right)+\left(x+1\right)\right]=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2+1}=a\left(a\ge0\right)\\\sqrt{x^2-x+1}=b\left(b>0\right)\end{cases}}\)

Tìm được \(\orbr{\begin{cases}a=2b\\b=2a\end{cases}}\)

TH1: a=2b => phương trình vô nghiệm

TH2: b=2a ta được \(x_1=\frac{5+\sqrt{37}}{2};x_2=\frac{5-\sqrt{37}}{2}\left(tmđk\right)\)

2x=5-3

2x=2

x=2:2

x=1

Vậy x=1

\(2x+3=5\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

2x+3=5

2x=5-3

2x=2

x=2:2

x=1

\(cotx=cot70^0\)

\(\Rightarrow x=70^0+k180^0\) (\(k\in Z\))

`|x-6|=-5x+9`     `ĐK: x <= 9/5`

`<=>[(x-6=-5x+9),(x-6=5x-9):}`

`<=>[(x=5/2 (ko t//m)),(x=3/4(t//m)):}`

 `|x - 6| = -5x + 9`

\(\Leftrightarrow\left[{}\begin{matrix}x-6=-5x+9\\x-6=5x-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5x=9+6\\x-5x=-9+6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=15\\-4x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{6}=\dfrac{5}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)

⇔⎡⎢ ⎢⎣x=156=52x=34

\(\Leftrightarrow2cos\frac{3x}{2}.cos\frac{x}{2}=2sin\frac{3x}{2}.cos\frac{3x}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{3x}{2}=0\\cos\frac{x}{2}=sin\frac{3x}{2}=cos\left(\frac{\pi}{2}-3x\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+\frac{k2\pi}{3}\\x=\frac{\pi}{7}+\frac{k4\pi}{7}\\x=\frac{\pi}{5}+\frac{k4\pi}{5}\end{matrix}\right.\)

\(\Leftrightarrow2\sin2x=\sqrt{2}\)

\(\Leftrightarrow\sin2x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\sin2x=\dfrac{\pi}{4}\)

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

it's not me OoO

\(\sin x=\dfrac{1}{2}\Leftrightarrow\sin x=\sin\dfrac{\pi}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)