Lời giải:

PT $\Leftrightarrow 4x^2+4x+1=3(x^2-4)+18$

$\Leftrightarrow 4x^2+4x+1=3x^2+6$

$\Leftrightarrow x^2+4x-5=0$

$\Leftrightarrow (x-1)(x+5)=0$

$\Leftrightarrow x-1=0$ hoặc $x+5=0$

$\Leftrightarrow x=1$ hoặc $x=-5$

\(\left(2x+1\right)^2=3\left(x-2\right)\left(x+2\right)+18\)

\(\Leftrightarrow4x^2+4x+1=3\left(x^2-4\right)+18\)

\(\Leftrightarrow4x^2+4x+1=3x^2-12+18\)

\(\Leftrightarrow4x^2+4x+1=3x^2+6\)

\(\Leftrightarrow4x^2-3x^2+4x=6-1\)

\(\Leftrightarrow x^2+4x=5\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow x^2+5x-x-5=0\)

\(\Leftrightarrow x\left(x+5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-5;1\right\}\)

\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)

\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)

\(\Leftrightarrow\left[4\left(x^2+2x\right)+3\right]\left(x^2+2x+1\right)-18=0\)

Đặt \(t=x^2+2x\)ta có

\(\left(4t+3\right)\left(t+1\right)-18=0\)

\(\Leftrightarrow4t^2+7x-15=0\)

\(\Leftrightarrow4t^2+12t-5t-15=0\)

\(\Leftrightarrow4t\left(t+3\right)-5\left(t+3\right)=0\)

\(\Leftrightarrow\left(t+3\right)\left(4t-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+3=0\\4t-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-3\\t=\frac{5}{4}\end{cases}}}\)

Nếu \(t=-3\Rightarrow x^2+2x=-3\)

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

\(\Rightarrow\)x vô nghiệm vì \(x^2+2x+3>0\)với mọi x

Nếu \(t=\frac{5}{4}\Rightarrow x^2+2x=\frac{5}{4}\)

\(\Leftrightarrow x^2+2x-\frac{5}{4}=0\)

\(\Leftrightarrow4x^2+8x-5=0\)

\(\Leftrightarrow4x^2-2x+10x-5=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}}\)

Vậy \(S=\left\{-\frac{5}{2};\frac{1}{2}\right\}\)

P/s tham khảo nha

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

2x-3=x+1/2

a,2x-3=x+1/2                       b,4x-(x+1/2)=2x+(1/2-5)                           c,2/3-1/3(x-2/3)-1/2(2x+1)=5

2x-x =1/2+3                           4x-x-1/2=2x+1/2-5                             d,(x+1/2).(x-3/4)=0

x=7/2                                4x-x-2x  =1/2-5+1/2                                 \(\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

                                            x=-4

e,(2x-1)(3x+1/5)=0

\(\orbr{\begin{cases}2x-1=0\\3x+\frac{1}{5}=0\end{cases}}\orbr{\begin{cases}2x=1\\3x=\frac{1}{5}\end{cases}}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{15}\end{cases}}\)

f, 4x²-2x=0

Các câu mk chưa làm thì bạn cứ chờ để mk suy nghĩ.

Đặt \(\left\{{}\begin{matrix}x-2y=a\\\dfrac{1}{2x+3y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=2\\2a+3b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=3\\\dfrac{1}{2x+3y}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2x+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2\left(3+2y\right)+3y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\6+4y+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3+2.-1\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Vậy nghiệm hpt \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

`3x^2-2x=x^2+3`

`<=>3x^2-x^2-2x-3=0`

`<=>2x^2-2x-3=0`

Ptr có: `\Delta'=(-1)^2-2.(-3)=7 > 0`

  `=>`Ptr có `2` nghiệm pb

`=>{(x_1=[-b'+\sqrt{\Delta'}]/a=[1+\sqrt{7}]/2),(x_1=[-b'-\sqrt{\Delta'}]/a=[1-\sqrt{7}]/2):}`