\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)

\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)

Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)

\(\Rightarrow4ab=\left(a+b\right)^2\)

\(\Leftrightarrow4ab=a^2+a^2+2ab\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow x^4-16x-12=0\)

\(\Leftrightarrow\left(x^2-2x-2\right)\left(x^2+2x+6\right)=0\)

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

\(\Rightarrow x=1\pm\sqrt{3}\)

Lời giải:

Ta có:

\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)

\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)

\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)

Đặt \(x^2-12x-48=a\). PT trở thành:

\((a-x)(a+20x)+20x^2=0\)

\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)

\(\Leftrightarrow a(a+19x)=0\)

\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)

\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)

Vậy......

\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)

\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)

\(x^4-8x^3+21x^2-24x+9=0\)

\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))

\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

Đk:\(x\ge1;x\le-2\)

Đặt \(t=\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}\)

\(\Rightarrow t^2=\left(x-1\right)\left(x+2\right)\)

Pttt: \(t^2+4t=12\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)

TH1: \(t=2\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-1\right)\left(x+2\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x^2+x-6=0\end{matrix}\right.\)\(\Rightarrow x=2\) (thỏa mãn)

TH2:\(t=-6\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=-6\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1< 0\\\left(x-1\right)\left(x+2\right)=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x^2+x-38=0\end{matrix}\right.\)\(\Rightarrow x=\dfrac{-1-3\sqrt{17}}{2}\) (thỏa mãn)

Vậy...

\(\left(x+2\right)\left(x-3\right)+3=\left(x-4\right)\left(x+2\right)-7\)

\(\Leftrightarrow x^2-x-6+3=x^2-2x-8-7\)

\(\Leftrightarrow x^2-x-x^2+2x=6-3-8-7\)

\(\Leftrightarrow x=-12\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-12\right\}\)

\(x^2-3x+2x-6+3-x^2-2x+4x+8+7=0\)

\(x+12=0\)

\(x=-12\)

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)

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

\(\Leftrightarrow t^2=sin^2+2sinx.cosx+cos^2x\)

\(\Leftrightarrow t^2=1+2sinx.cosx\)

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

Pttt: \(3t-4.\dfrac{t^2-1}{2}=0\) \(\Leftrightarrow-2t^2+3t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(ktm\right)\\t=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow sinx.cosx=-\dfrac{3}{8}\) \(\Leftrightarrow2sinx.cosx=-\dfrac{3}{4}\)\(\Leftrightarrow sin2x=-\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\end{matrix}\right.\), \(k\in Z\)

Vậy...

b)Pt 

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

\(\Leftrightarrow t^2-1=-2sinx.cosx\)

Pttt:\(12t+t^2-1=2\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-6+\sqrt{39}\left(tm\right)\\t=-6-\sqrt{39}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow cosx+sinx=-6+\sqrt{39}\)

\(\Leftrightarrow\sqrt{2}.cos\left(x-\dfrac{\pi}{4}\right)=-6+\sqrt{39}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arc.cos\left(\dfrac{-6+\sqrt{39}}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{\pi}{4}-arc.cos\left(\dfrac{-6+\sqrt{39}}{2}\right)+k2\pi\end{matrix}\right.\)\(,k\in Z\)

Vậy...(Nghiệm xấu)

Lần sau đừng tự tiện xếp vào phần bất pt bạn nhé :(

Ta có : \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)=3x^2\)

\(\Leftrightarrow4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)=3x^2\)

\(\Leftrightarrow4\left(x^2+17x+60\right)\left(x^2+16x+60\right)=3x^2\)(1)

Đặt \(x^2+16x+60=a\)

Pt (1) \(\Leftrightarrow4\left(a+x\right)a=3x^2\)

         \(\Leftrightarrow4\left(a^2+ax\right)=3x^2\)

          \(\Leftrightarrow4a^2+4ax=3x^2\)

          \(\Leftrightarrow4a^2+4ax+x^2=4x^2\)

         \(\Leftrightarrow\left(2a+x\right)^2=4x^2\)

          \(\Leftrightarrow\orbr{\begin{cases}2a+x=2x\\2a+x=-2x\end{cases}}\)

*Nếu \(2a+x=2x\)

\(\Leftrightarrow2a=x\)

\(\Leftrightarrow x^2+16x+60=x\)

\(\Leftrightarrow x^2+15x+60=0\)

\(\Leftrightarrow x^2+2.\frac{15}{2}.x+\frac{225}{4}+\frac{15}{4}=0\)

\(\Leftrightarrow\left(x+\frac{15}{2}\right)^2+\frac{15}{4}=0\)

Pt vô nghiệm

*Nếu \(2a+x=-2x\)

\(\Leftrightarrow2a+3x=0\)

\(\Leftrightarrow2\left(x^2-16x+60\right)+3x=0\)

\(\Leftrightarrow2x^2-32x+120+3x=0\)

\(\Leftrightarrow2x^2-29x+120=0\)

\(\Leftrightarrow x^2-\frac{29}{2}x+60=0\)

\(\Leftrightarrow x^2-2.\frac{29}{4}.x+\frac{841}{16}+\frac{119}{16}=0\)

\(\Leftrightarrow\left(x-\frac{29}{4}\right)^2+\frac{119}{16}=0\)

Pt vô nghiệm

Vậy pt vô nghiệm