b.x2+ x - 1 = (x + 2)√(x2 - 2x + 2)
Đk:\(x\in\left(-\infty;\infty\right)\)
Bình 2 vế lên đc:
\(\left(x^2+x-1\right)^2=\left(x+2\right)^2\sqrt{\left(x^2-2x+2\right)^2}\)
\(\Leftrightarrow x^4+2x^3-x^2-2x+1=\left(x^2+4x+4\right)\left(x^2-2x+2\right)\)
\(\Leftrightarrow x^4+2x^3-x^2-2x+1=x^4+2x^3-2x^2+8\)
\(\Leftrightarrow x^2-2x-7=0\)
Theo Delta ta có:\(\Delta=\left(-2\right)^2-\left(-4\left(1.7\right)\right)=32>0\)
=>pt có 2 nghiệm thực
\(x_{1,2}=\frac{2\pm\sqrt{32}}{2}\)
a) ĐK: x\(\ge\frac{2}{3}\)
pt \(\Leftrightarrow\sqrt{4x+1}-37+\sqrt{3x-2}-32=\frac{x+3}{5}-69\)
\(\Leftrightarrow\frac{4x+1-1369}{\sqrt{4x+1}+37}+\frac{3x-2-1024}{\sqrt{3x-2}+32}-\frac{x+3}{5}+69=0\)
\(\Leftrightarrow\frac{4\left(x-342\right)}{\sqrt{4x+1}+37}+\frac{3\left(x-342\right)}{\sqrt{3x-2}+32}-\frac{x-342}{5}=0\)
\(\Leftrightarrow\left(x-342\right)\left(\frac{4}{\sqrt{4x+1}+37}+\frac{3}{\sqrt{3x-2}+32}-\frac{1}{5}\right)=0\left(#\right)\)
Với mọi \(x\ge\frac{2}{3}\) ta có:
\(\frac{4}{\sqrt{4x+1}+37}>0;\frac{3}{\sqrt{3x-2}+32}>0;\frac{1}{-5}< 0\);
\(\Rightarrow\frac{4}{\sqrt{4x+1}+37}+\frac{3}{\sqrt{3x-2}+32}-\frac{1}{5}\ne0\)
Do đó pt (#) \(\Leftrightarrow x-342=0\)
\(\Leftrightarrow x=342\)
