1.
[tex](2x+7)\sqrt{2x+7}=x^{2}+9x+7 \\\Leftrightarrow (2x+7)\sqrt{2x+7}=x^{2}+7x+(2x+7) \\\Leftrightarrow x^2+2x+7-2x\sqrt{2x+7}+7x-7.\sqrt{2x+7}=0\\\rightarrow (x-\sqrt{2x+7})^2-7(x-\sqrt{2x+7})=0[/tex]
Đặt [tex]x-\sqrt{2x+7}=t[/tex] có:
[tex]t^2-7t=0\rightarrow t(t-7)=0[/tex] chị lo nốt giúp em
2.
[tex]\sqrt{(2x^2 + 5x + 12)} + \sqrt{(2x^2 + 3x + 2)} = x + 5[/tex]
Đặt [tex]\sqrt{(2x^2 + 5x + 12)}=a \rightarrow 2x^2 + 5x + 12=a^2[/tex]
[tex]\sqrt{(2x^2 + 3x + 2)}=b\rightarrow 2x^2 + 3x + 2=b^2[/tex]
[tex]\rightarrow a^2-b^2=2x+10=2(x+5)[/tex]
mà [tex]a+b=x+5[/tex]
[tex]\rightarrow a^2-b^2=2(a+b)\rightarrow (a+b)(a-b)-2(a+b)=0\rightarrow (a+b)(a-b-2)=0[/tex]
Giải típ nhé chị!