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a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
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dk \(\hept{\begin{cases}3x^2-1\ge0\\x^2-x\ge0\end{cases}< =>\orbr{\begin{cases}x\ge1\\x\le\frac{-1}{\sqrt{3}}\end{cases}}}\)(1)
\(< =>2\sqrt{6x^2-2}+2\sqrt{2x^2-2x}-2x\sqrt{2x^2+2}\)=7x2-x+4
<=> (3x2-1)-2\(\sqrt{2}.\sqrt{3x^2-1}\)+ 2 + (x2+1)+2x\(\sqrt{2}.\sqrt{x^2+1}\)+2x2 + (x2-x) - 2\(\sqrt{2}\sqrt{x^2-x}\)+2 =0
<=> \(\left(\sqrt{3x^2-1}-1\right)^2+\left(\sqrt{x^2+1}+x\sqrt{2}\right)^2\)+\(\left(\sqrt{x^2-x}-\sqrt{2}\right)^2=0\)
<=> \(\hept{\begin{cases}\sqrt{3x^2-1}=\sqrt{2}\\\sqrt{x^2+1}+x\sqrt{2}=0\\\sqrt{x^2-x}=\sqrt{2}\end{cases}}< =>\hept{\begin{cases}3x^2=3\\x^2+1=2x^2\left(x< 0\right)\\x^2-x-2=0\end{cases}}\)<=> \(\hept{\begin{cases}x^2=1\\\left(x+1\right)\left(x-2\right)=0\end{cases}< =>x=-1}\) (thỏa mãn điều kiện (1)
vậy x=-1 là nghiệm
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dk \(x+9\ge0;x\ge0;x+1>0< =>x\ge0;\)
\(\sqrt{x+9}-\sqrt{x}=\frac{2\sqrt{2}}{\sqrt{x+1}}< =>\frac{9}{\sqrt{x+9}+\sqrt{x}}=\frac{2\sqrt{2}}{\sqrt{x+1}}\)<=> \(9\sqrt{x+1}=2\sqrt{2}\left(\sqrt{x+9}+\sqrt{x}\right)< =>\)\(81\left(x+1\right)=16x+72+16\sqrt{x\left(x+9\right)}\)
<=> \(65x+9=16\sqrt{x\left(x+9\right)}\)<=> 4225x2+1170x+81= 256x2+144x <=> 3969x2+1026x+81=0 (vô nghiệm)
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dk \(\hept{\begin{cases}x\left(3x+1\right)\ge0\\x\left(x-1\right)\ge0\end{cases}< =>\orbr{\begin{cases}x\ge1\\x\le\frac{-1}{3}\end{cases}}}\)
vì x khác 0 nên chia cả 2 vế cho \(\sqrt{x}\)ta được \(\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{x}< =>\)\(\sqrt{x-1}+2\sqrt{x}-\sqrt{3x+1}=0< =>\)\(\sqrt{x-1}+\frac{4x-\left(3x+1\right)}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(\sqrt{x-1}+\frac{x-1}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(< =>\sqrt{x-1}\left(1+\frac{\sqrt{x-1}}{2\sqrt{x}+\sqrt{3x+1}}\right)=0< =>\sqrt{x-1}=0\) (vì biểu thức trong ngoặc luôn \(\ge1\)) <=> x-1= 0 <=> x=1 (thỏa mãn điều kiện)
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Lời giải:
ĐKXĐ: \(x^2\geq 5\)
PT \(\Leftrightarrow (\sqrt{x^2+7}-4)-(\sqrt{x^2-5}-2)=x-3\)
\(\Leftrightarrow \frac{x^2+7-16}{\sqrt{x^2+7}+4}-\frac{x^2-5-4}{\sqrt{x^2-5}+2}=x-3\)
\(\Leftrightarrow \frac{(x-3)(x+3)}{\sqrt{x^2+7}+4}-\frac{(x-3)(x+3)}{\sqrt{x^2-5}+2}=x-3\)
\(\Leftrightarrow (x-3)\left[1+\frac{x+3}{\sqrt{x^2-5}+2}-\frac{x+3}{\sqrt{x^2+7}+4}\right]=0(1)\)
Với \(\forall x^2\geq 5\) thì:
\(\left\{\begin{matrix} x+3>0\\ \sqrt{x^2-5}+2< \sqrt{x^2+7}+4\end{matrix}\right.\Rightarrow \frac{x+3}{\sqrt{x^2-5}+2}>\frac{x+3}{\sqrt{x^2+7}+4}\)
\(\Rightarrow 1+\frac{x+3}{\sqrt{x^2-5}+2}-\frac{x+3}{\sqrt{x^2+7}+4}\neq 0(2)\)
Từ (1);(2) \(\Rightarrow x-3=0\Rightarrow x=3\) (thỏa mãn)
Vậy.......
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ĐKXĐ: ...
\(\Leftrightarrow3\left(2\sqrt{x+2}+\sqrt{3-x}\right)=3x+1+4\sqrt{-x^2+x+6}\)
Đặt \(2\sqrt{x+2}+\sqrt{3-x}=t>0\)
\(\Rightarrow t^2=4\left(x+2\right)+3-x+4\sqrt{\left(x+2\right)\left(3-x\right)}=3x+11+4\sqrt{-x^2+x+6}\)
Pt trở thành:
\(3t=t^2-10\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2\sqrt{x+2}+\sqrt{3-x}=5\)
Ta có: \(VT=2\sqrt{x+2}+\sqrt{3-x}\le\sqrt{\left(2^2+1^2\right)\left(x+2+3-x\right)}=5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\frac{\sqrt{x+2}}{2}=\sqrt{3-x}\Leftrightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)