sqrt{x^2+8}-7xsqrt{x^2+3}-6(1)Leftrightarrowsqrt{x^2+8}-37x-7+sqrt{x^2+3}-2Leftrightarrowfrac{left(sqrt{x^2+8}-3right)left(sqrt{x^2+8}+3right)}{left(sqrt{x^2+8}+3right)}7left(x-1right)+frac{left(sqrt{x^2+3}-2right)left(sqrt{x^2+3}+2right)}{sqrt{x^2+3}+2}Leftrightarrowfrac{x^2+8-9}{left(sqrt{x^2+8}+3right)}7left(x-1right)+frac{x^2-1}{sqrt{x^2+3}+2}Leftrightarrowfrac{x^2-1}{sqrt{x^2+8}+3}-7left(x-1right)-frac{x^2-1}{sqrt{x^2+3+2}}0Leftrightarrowleft(x-1right)left(frac{x+1}{sqrt{x^2+8}+3}-7-fra...
Đọc tiếp
\(\sqrt{x^2+8}-7x=\sqrt{x^2+3}-6\)(1)
\(\Leftrightarrow\sqrt{x^2+8}-3=7x-7+\sqrt{x^2+3}-2\)
\(\Leftrightarrow\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{\left(\sqrt{x^2+3}-2\right)\left(\sqrt{x^2+3}+2\right)}{\sqrt{x^2+3}+2}\)
\(\Leftrightarrow\frac{x^2+8-9}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+3}+2}\)
\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+8}+3}-7\left(x-1\right)-\frac{x^2-1}{\sqrt{x^2+3+2}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}\right)=0\)
\(\Leftrightarrow x-1=0\)
hay \(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}=0\)(2)
Từ (1), có:
\(\sqrt{x^2+8}-\sqrt{x^2+3}=7x-6>0\)
\(\Leftrightarrow7x-6>0\)
\(\Leftrightarrow x>\frac{6}{7}\)
Khi đó, có:
\(\frac{x+1}{\sqrt{x^2+8}+3}-\frac{\sqrt{x+1}}{\sqrt{x^2+3}+2}<0\)
\(\Rightarrow\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+3}+2}-7<0\)
Vậy, pt (2) vô nghiệm
Do đó, pt (1) có 1 nghiệm là x = 1
