b. \(\sqrt{x-4}+\sqrt{x^2-3x+4}=x\)
(ĐKXĐ: \(x\ge4\))
\(\Leftrightarrow\sqrt{x^2-3x+4}=x-\sqrt{x-4}\)
\(\Leftrightarrow x^2-3x+4=x^2+x-4-2\sqrt{x\left(x-4\right)}\)
\(\Leftrightarrow x^2-3x+4-x^2-x+4+2\sqrt{x^2-4x}=0\Leftrightarrow-4x+8+2\sqrt{x^2-4x}=0\Leftrightarrow-2\left(2x-4-\sqrt{x^2-4x}\right)=0\Leftrightarrow2x-4-\sqrt{x^2-4x}=0\Leftrightarrow\sqrt{x^2-4x}=2x-4\Leftrightarrow x^2-4x=4x^2+16-16x\Leftrightarrow x^2-4x^2-4x+16x-16=0\Leftrightarrow-3x^2+12x-16=0\Leftrightarrow3x^2-12x+16=0\)
Ta có: \(\Delta=b^2-4ac=\left(-12\right)^2-4.3.16=-48< 0\)
=> pt vô nghiệm.
Vậy pt đã cho vô nghiệm.
ĐK : x > 3/2
Đặt \(\sqrt{3x-2}=a\left(a>0\right)\) . Khi đó pt thành :
\(1+\dfrac{x}{a}=\dfrac{1+a}{x}\Leftrightarrow\dfrac{a+x}{a}=\dfrac{a+1}{x}\Leftrightarrow a^2+a=ax+x^2\Leftrightarrow x^2+a\left(x-1\right)-a^2=0\)
hay \(\sqrt{3x-2}\left(x-1\right)+x^2-3x+2=0\Leftrightarrow\left(\sqrt{3x-2}-1\right)\left(x-1\right)+x^2-2x+1=0\Leftrightarrow\dfrac{3x-3}{\sqrt{3x-2}+1}\left(x-1\right)+\left(x-1\right)^2=0\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x-2}+1}+\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x-2}+1}+1\right)=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
Vì \(\dfrac{3}{\sqrt{3x-2}+1}+1>0\)
Vậy nghiệm của pt là x = 1
