a/ \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
b/ \(\left(x-2\right)^3+\left(5-2x\right)^3=0\)
\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)
\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4+2x^2-5x-4x+10+25-20x+4x^2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2-33x+29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\7x^2-33x+29=0\end{matrix}\right.\)
+) 3 - x = 0 => x = 3
+) \(7x^2-33x+29=0\)
\(\Leftrightarrow\left(\sqrt{7}x\right)^2-2\cdot\sqrt{7}x\cdot\dfrac{33\sqrt{7}}{14}+\dfrac{1089}{28}-\dfrac{277}{28}=0\)
\(\Leftrightarrow\left(\sqrt{7}x-\dfrac{33\sqrt{7}}{14}\right)^2=\dfrac{277}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{7}x-\dfrac{33\sqrt{7}}{14}=\sqrt{\dfrac{277}{28}}\\\sqrt{7}x-\dfrac{33\sqrt{7}}{28}=-\sqrt{\dfrac{277}{28}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33+\sqrt{277}}{14}\\x=\dfrac{33-\sqrt{277}}{14}\end{matrix}\right.\)
Thử lại thấy chỉ có x = 3 thỏa mãn
Vậy pt có 1 nghiệm x = 3
bài 2:
a/ \(\left(n+3\right)^2-\left(n-1\right)^2=\left(n+3-n+1\right)\left(n+3+n-1\right)=4\left(2n+2\right)=8\left(n+1\right)⋮8\left(đpcm\right)\)
b/ \(\left(n+6\right)^2-\left(n-6\right)^2=\left(n+6-n+6\right)\left(n+6+n-6\right)=12\cdot2n=24n⋮24\)