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a)\(pt\Leftrightarrow\sqrt{x^2-2x+2}+\sqrt{3x^2-6x+4}-2=0\)
\(\Leftrightarrow\sqrt{x^2-2x+2}-1+\sqrt{3x^2-6x+4}-1=0\)
\(\Leftrightarrow\frac{x^2-2x+2-1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\frac{x^2-2x+1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+3}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{x^2-2x+2}+1}+\frac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}>0\) (loại)
Nên x-1=0 suy ra x=1
b)\(pt\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}+x^2+2x-5=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+21}-4+x^2+2x+1=0\)
\(\Leftrightarrow\frac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\frac{5x^2+10x+21-16}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)
\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\frac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1\right)=0\)
Dễ thấY: \(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1>0\) (loại luôn)
Nên x+1=0 suy ra x=-1
2) pt đề bài cho=0
<=> \(\left(x-1\right)^2\left(2x^2-x+2\right)\)=0
<=>\(\orbr{\begin{cases}x-1=0\left(1\right)\\2x^2-x+2=0\left(2\right)\end{cases}}\)
Từ 1 => x=1
từ 2 =>\(2\left(x^2-\frac{1}{2}x+1\right)\)
=\(2\left[\left(x-\frac{1}{4}\right)^2+\frac{15}{16}\right]>0\)với mọi x
Nên pt 2 cô nghiệm
Vậy pt đề cho có nghiệm là 1
\(\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\\ =\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)\cdot\left(3\sqrt{x}+14\right)}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+14}{5\sqrt{x}-1}\)