Question

Giải các phương tình sau:

a) \(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

b)\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)

c)\(2x\left(8x-1\right)^2\left(4x-1\right)=0\)

d)\(x^2-y^2+2x-4y-10=0\) ( x,y là các số nguyên dương )

Answer 1

d)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)=7\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)

Mà x,y nguyên dương\(\Rightarrow x-y-1< x+y+3\)

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

Answer 2

Mạn phép ko chép lại đề :

b) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)

⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0)

⇔ \(8\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)

⇔ ( x + 4)² = 16

⇔ x² + 8x + 16 = 16

⇔ x( x + 8) = 0

⇔ x = 0 ( KTM) hoặc : x = - 8 ( TM)

KL.....