MÌnh nghĩ là bình phương 2 vế lên. CÁch làm như sau:

\(\left(\left(x+3\right)\sqrt{\left(4-x\right)\left(12+x\right)}\right)^2=\left(28-x\right)^2\)

Chắc bạn đã học (axb)²=a²x b². ÁP dụng vào thôi:

=>(x+3)² (4-x)(12+x) = (28-x)²

=>(x²+6x+9)(48-8x-x²)=784-56x+x²

=>48x²+288x+432-8x³-48x-72x-x⁴-6x³-9x²=784-56x+x²

=>39x²+168x+432-14x³-x⁴=784-56+x²

=>-x⁴-14x³+38x²+168x-296=0

đến đó bạn thử giải XEM

Xin lỗi vì đã không thể giúp bạn. chúc bạn luôn học tốt

\(\left(x+3\right)\sqrt{\left(4-x\right).\left(12+x\right)}=28-x\\ \Leftrightarrow\left(x+3\right)\sqrt{48+4x-12x-x^2}=28-x\\ \Leftrightarrow\left(x+3\right)\sqrt{-x^2-8x+48}=28-x\\ \Leftrightarrow\\ \left[\left(x+3\right)\sqrt{-x^2-8x+48}\right]^2=\left(28-x\right)^2\\ \Leftrightarrow\left(x+3\right)^2\left(-x^2-8x+48\right)=784-56x+x^2\\ \Leftrightarrow-\left(x^2+6x+9\right)\left(x^2+8x+48\right)=784-56x+x^2\\ \Leftrightarrow-\left(x^4+8x^3+48x^2+6x^3+48x^2+288x+9x^2+72x+432\right)=784-56x+x^2\\ \Leftrightarrow-x^4-14x^3-105x^2-360x-432-784+56x-x^2=0\\ \Leftrightarrow-x^4-14x^3-107x^2-416x-1216=0\)

Mình làm tới bước này rồi, cậu có thể nhờ máy tính giải hộ ạ

Bài 1

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...

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.

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

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

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

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

\(\Leftrightarrow...\)

mong mọi người giải giúp em vs 

ĐKXĐ: \(-4\le x\le1\)

Đặt \(\sqrt{x+4}-\sqrt{1-x}=t\)

\(\Rightarrow t^2=5-2\sqrt{\left(x+4\right)\left(1-x\right)}\Rightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=\frac{5-t^2}{2}\)

Pt trở thành:

\(t\left(1+\frac{5-t^2}{2}\right)=3\Leftrightarrow t\left(7-t^2\right)=6\)

\(\Leftrightarrow t^3-7t+6=0\Leftrightarrow\left(t+3\right)\left(t-1\right)\left(t-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=1\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x+4}-\sqrt{1-x}=-3\\\sqrt{x+4}-\sqrt{1-x}=1\\\sqrt{x+4}-\sqrt{1-x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+4}+3=\sqrt{1-x}\left(vn\right)\\\sqrt{x+4}=1+\sqrt{1-x}\\\sqrt{x+4}=2+\sqrt{1-x}\end{matrix}\right.\) (1 vô nghiệm do \(VT\ge3;VP\le\sqrt{5}< 3\))

\(\Leftrightarrow\left[{}\begin{matrix}x+4=2-x+2\sqrt{1-x}\\x+4=5-x+4\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{1-x}\left(x\ge-1\right)\\2x-1=4\sqrt{1-x}\left(x\ge\frac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=1-x\\4x^2-4x+1=16-16x\end{matrix}\right.\) \(\Leftrightarrow...\)