Câu 1:
Đặt \(3x-16y-24=k\left(k\in N\right)\) khi đó:
\(\sqrt{9x^2+16x+32}=k\Rightarrow9x^2+16x+32=k^2\)
\(\Rightarrow9\left(x+\dfrac{8}{9}\right)^2+\dfrac{224}{9}=k^2\)
\(\Rightarrow\dfrac{1}{9}\left(\left(9x+8\right)^2-9k^2\right)=-\dfrac{224}{9}\)
\(\Rightarrow\left(9x+8+3k\right)\left(9x+8-3k\right)=-224\)
tự giải nốt
Câu 2:
\(4x^3+5x^2+1=\sqrt{3x+1}-3x\)
\(\Leftrightarrow4x^3+5x^2+3x+1=\sqrt{3x+1}\)
\(\Leftrightarrow 16x^6+40x^5+49x^4+38x^3+19x^2+6x+1=3x+1\)
\(\Leftrightarow x(4x+1)(4x^4+9x^3+10x^2+7x+3)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(3x-y-24=\sqrt{9x^2+16+32}\) (1)
\(x,y\in Z\Rightarrow9x^2+16x+32=k^2\Rightarrow64-9.32+9k^2=t^2\)
\(\left(3k\right)^2-t^2=32.7=2^5.7\)
\(\Rightarrow\left(I\right)\left\{{}\begin{matrix}3k-t=14\\3k+t=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=5\\t=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{9}\left(loai\right)\end{matrix}\right.\)
(1)\(\Rightarrow3.\left(-1\right)-y-24=25\Rightarrow y=-52\)
\(\left(x,y\right)=\left(-1,-52\right)\)
có thể còn nhiều nữa tự làm
