\(x^4-8x^3+22x^2-24x+7+2m=0\)

\(\Leftrightarrow\left(x^2-4x\right)^2+6\left(x^2-4x\right)+7+2m=0\)

\(đặt:x^2-4x=t\Rightarrow t^2+6t+7+2m=0\)

\(\Delta=8-8m>0\Leftrightarrow m< 1\Rightarrow\left[{}\begin{matrix}t1=\dfrac{-6+\sqrt{8-8m}}{2}=\sqrt{2-2m}-3\\t2=\dfrac{-6-\sqrt{8-8m}}{2}=-\sqrt{2-2m}-3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2-4x-\sqrt{2-2m}+3=0\left(1\right)\\x^2-4x+\sqrt{2-2m}+3=0\left(2\right)\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\left(2\right)có-2-ngo-pb-và-không-có-nghiệm-chung\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'1>0\Leftrightarrow1+\sqrt{2-2m}>0\\\Delta'2>0\Leftrightarrow1-\sqrt{2-2m}>0\end{matrix}\right.\)\(\Leftrightarrow\dfrac{1}{2}< m\le1\left(m< 1\right)\Rightarrow\dfrac{1}{2}< m< 1\left(3\right)\)

\(giả-sử-có-ngo-chung\Rightarrow\left(1\right)-\left(2\right)=0\)

\(\Rightarrow-2\sqrt{2-2m}=0\Leftrightarrow m=1\Rightarrow m\ne1\left(4\right)\)

\(\left(3\right)\left(4\right)\Rightarrow\dfrac{1}{2}< m< 1\)

\(\Leftrightarrow\left(sin4x-cos4x\right)\left(sin4x+4cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x-cos4x=0\\sin4x=-4cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(4x-\frac{\pi}{4}\right)=0\\tan4x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{4}=k\pi\\4x=arctan\left(-4\right)+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{1}{4}arctan\left(-4\right)+\frac{k\pi}{4}\end{matrix}\right.\)

Pt có 4 nghiệm trên khoảng đã cho

Ta có:

\(7x^2+y^2+4xy-24x-6y+21=0\)

\(\Leftrightarrow y^2+4xy-6y+7x^2-24x+21=0\)

\(\Leftrightarrow y^2+2y\left(2x-3\right)+\left(2x-3\right)^2+3x^2-12x+12=0\)

\(\Leftrightarrow\left(y+2x-3\right)^2+3\left(x^2-4x+4\right)=0\)

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

Mà \(\hept{\begin{cases}\left(y+2x-3\right)^2\ge0\\3\left(x-2\right)^2\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+2x-3=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)

Vậy cặp số  \(\left(x,y\right)=\left(2;-1\right)\)

mink vẫn chưa hiểu lắm bn ak giảng lại cho mink hiểu đi

Đề yêu cầu gì em?

\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)

\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(2x^2-3x+2=2\left(x+\dfrac{1}{2}\right)\left(x-2\right)\)

\(4x^2-7x-2=4\left(x-2\right)\left(x+\dfrac{1}{4}\right)\)

\(4x^2+15x+9=4\left(x+\dfrac{3}{4}\right)\left(x+3\right)\)

\(\Leftrightarrow x^4-x^3+x^3-x^2-8x^2+8x+16x-16=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+4x^2-3x^2-12x+4x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x^2-3x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\\\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\left(vô.n_o\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)