172a.
ĐKXĐ: \(0\le x\le1\)
\(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)
\(\Leftrightarrow\sqrt{x+\sqrt{1-x}}=1-\sqrt{x}\)
\(\Rightarrow x+\sqrt{1-x}=\left(1-\sqrt{x}\right)^2\)
\(\Rightarrow x+\sqrt{1-x}=1-2\sqrt{x}+\sqrt{x}\)
\(\Rightarrow\sqrt{1-x}=1-2\sqrt{x}\)
\(\Rightarrow\left\{{}\begin{matrix}1-2\sqrt{x}\ge0\\1-x=\left(1-2\sqrt{x}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\1-x=1-4\sqrt{x}+4x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\x\left(5\sqrt{x}-4\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{16}{25}>\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)
Vậy \(x=0\)
172b.
ĐKXĐ: \(\left[{}\begin{matrix}x=0\\x\ge1\end{matrix}\right.\)
- Với \(x=0\) ko phải nghiệm
- Với \(x\ge1\)
\(\sqrt{1-\sqrt{x^2-x}}=\sqrt{x}-1\)
\(\Leftrightarrow1-\sqrt{x^2-x}=x-2\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x^2-x}=2\sqrt{x}-x\)
\(\Leftrightarrow\sqrt{x-1}=2-\sqrt{x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2-\sqrt{x}\ge0\\x-1=\left(2-\sqrt{x}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x-1=4+x-4\sqrt{x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\\sqrt{x}=\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{25}{16}\)
172c.
ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
\(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
Ta có: \(\sqrt{x^2+6}>\sqrt{x^2}=\left|x\right|\ge x\) (1)
\(2\sqrt{x^2-1}\ge0\Rightarrow x-2\sqrt{x^2-1}\le x\) (2)
(1);(2) \(\Rightarrow\sqrt{x^2+6}>x-2\sqrt{x^2-1}\)
Pt vô nghiệm
d.
\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x=-1\) là 1 nghiệm
- Với \(x< -1\Rightarrow\) vế trái dương vế phải âm nên pt vô nghiệm
- Với \(x>-1\Rightarrow x>1\) (ĐKXĐ) pt tương đương:
\(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2\left(x+3\right)+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\)
\(\Leftrightarrow8\left(x+3\right)\left(x-1\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\8\left(x+3\right)=x-1\Rightarrow x=-\dfrac{25}{7}< 1\left(loại\right)\end{matrix}\right.\)
172e.
ĐKXĐ: \(7\le x\le9\)
Ta có:
\(VT=\sqrt{x-7}+\sqrt{9-x}\le\sqrt{2\left(x-7+9-x\right)}=2\)
\(VP=\left(x-8\right)^2+2\ge2\)
\(\Rightarrow VT\le VP\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-7}=\sqrt{9-x}\\x-8=0\end{matrix}\right.\)
\(\Rightarrow x=8\)
173a.
ĐKXĐ: \(x\ge2\)
\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
\(\Leftrightarrow3x-3+2\sqrt{\left(2x-1\right)\left(x-2\right)}=x+1\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-2\right)}=2-x\) (1)
\(\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\\left(2x-1\right)\left(x-2\right)=\left(2-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\2x^2-5x+2=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x^2-x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1< 2\left(loại\right)\\x=2\end{matrix}\right.\) (nhận)
Hoặc là ko cần giải như vậy, tới bước (1) chỉ cần biện luận:
- Với \(x=2\) thỏa mãn
- Với \(x>2\) vế trái dương, vế phải âm nên pt vô nghiệm
Vậy \(x=2\) là nghiệm duy nhất
173b.
ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow\sqrt{3x+15}=\sqrt{4x+17}+\sqrt{x+2}\)
\(\Leftrightarrow3x+15=5x+19+2\sqrt{\left(4x+17\right)\left(x+2\right)}\)
\(\Leftrightarrow\sqrt{\left(4x+17\right)\left(x+2\right)}+x+2=0\)
- Với \(x=-2\) là 1 nghiệm
Với \(x>-2\Rightarrow\left\{{}\begin{matrix}\sqrt{\left(4x+17\right)\left(x+2\right)}>0\\x+2>0\end{matrix}\right.\)
\(\Rightarrow\sqrt{\left(4x+17\right)\left(x+2\right)}+x+2>0\Rightarrow\) pt vô nghiệm
Vậy pt có nghiệm duy nhất \(x=-2\)
173c.
ĐKXĐ: \(x\ge1\)
Do \(x\ge1\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}\ge0\\\sqrt{x+3}\ge\sqrt{1+3}=2\end{matrix}\right.\)
\(\Rightarrow VT\ge2\)
Đồng thời \(VP=4-2x\le4-2.2=2\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra khi và chỉ khi \(x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
173d.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2x+11+2\sqrt{\left(x+1\right)\left(x+10\right)}=2x+7+2\sqrt{\left(x+2\right)\left(x+5\right)}\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+10\right)}+2=\sqrt{\left(x+2\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+10\right)+4\sqrt{\left(x+1\right)\left(x+10\right)}+4=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+10\right)}+x+1=0\)
- Với \(x=-1\) là 1 nghiệm
- Với \(x>-1\Rightarrow\left\{{}\begin{matrix}4\sqrt{\left(x+1\right)\left(x+10\right)}>0\\x+1>0\end{matrix}\right.\)
\(\Rightarrow4\sqrt{\left(x+1\right)\left(x+10\right)}+x+1>0\Rightarrow\) pt vô nghiệm
Vậy pt có nghiệm duy nhất \(x=-1\)
174a.
ĐKXĐ: \(x\ge2\)
Nhân 2 vế của pt với \(\sqrt{2x+3+\sqrt{x+2}}-\sqrt{2x+2-\sqrt{x-2}}\) rồi rút gọn ta được:
\(2x+3+\sqrt{x+2}-\left(2x+2-\sqrt{x+2}\right)=\left(1+2\sqrt{x+2}\right)\left(\sqrt{2x+3+\sqrt{x+2}}-\sqrt{2x+2-\sqrt{x+2}}\right)\)
\(\Leftrightarrow1+2\sqrt{x+2}=\left(1+2\sqrt{x+2}\right)\left(\sqrt{2x+3+\sqrt{x+2}}-\sqrt{2x+2-\sqrt{x+2}}\right)\)
\(\Leftrightarrow\sqrt{2x+3+\sqrt{x+2}}-\sqrt{2x+2-\sqrt{x+2}}=1\)
\(\Leftrightarrow\sqrt{2x+3+\sqrt{x+2}}=1+\sqrt{2x+2-\sqrt{x+2}}\)
\(\Leftrightarrow2x+3+\sqrt{x+2}=1+2x+2-\sqrt{x+2}+2\sqrt{2x+2-\sqrt{x+2}}\)
\(\Leftrightarrow\sqrt{x+2}=\sqrt{2x+2-\sqrt{x+2}}\)
\(\Leftrightarrow x+2=2x+2-\sqrt{x+2}\)
\(\Leftrightarrow\sqrt{x+2}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+2=x^2\end{matrix}\right.\)
\(\Rightarrow x=2\)
174b.
ĐKXĐ: \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x\ge4\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-4\right)}+3\sqrt{2x-1}=\sqrt{\left(2x-1\right)\left(x+11\right)}\)
- Với \(x=\dfrac{1}{2}\) là 1 nghiệm
- Với \(x\ne\dfrac{1}{2}\Rightarrow x\ge4\) pt tương đương:
\(\sqrt{x-4}+3=\sqrt{x+11}\)
\(\Leftrightarrow x+5+6\sqrt{x-4}=x+11\)
\(\Leftrightarrow\sqrt{x-4}=1\)
\(\Leftrightarrow x=5\)
175a.
ĐKXĐ: \(x\le1\)
Do \(x\le1\Rightarrow x-2< 0\Rightarrow x-2=-\sqrt{\left(x-2\right)^2}\)
Pt trở thành:
\(\sqrt{1-x}+\sqrt{\left(2-x\right)\left(1-x\right)}-\sqrt{\dfrac{\left(x-2\right)^2\left(x-1\right)}{x-2}}=3\)
\(\Leftrightarrow\sqrt{1-x}+\sqrt{\left(1-x\right)\left(2-x\right)}-\sqrt{\left(1-x\right)\left(2-x\right)}=3\)
\(\Leftrightarrow\sqrt{1-x}=3\)
\(\Leftrightarrow1-x=9\)
\(\Leftrightarrow x=-8\)
175b.
ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-2\end{matrix}\right.\)
Đặt \(t=\left(x-2\right)\sqrt{\dfrac{x+2}{x-2}}\)
\(\Rightarrow t^2=\dfrac{\left(x-2\right)^2\left(x+2\right)}{x-2}=\left(x-2\right)\left(x+2\right)\)
Pt trở thành:
\(t^2+4t=-3\Rightarrow t^2+4t+3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=-3\end{matrix}\right.\)
Do \(t< 0\Rightarrow\left(x-2\right)\sqrt{\dfrac{x+2}{x-2}}< 0\Rightarrow x< 2\)
\(\Rightarrow x\le-2\)
Khi đó:
- Với \(t=-1\Rightarrow t^2=1\Rightarrow\left(x-2\right)\left(x+2\right)=1\)
\(\Rightarrow x^2=5\Rightarrow x=-\sqrt{5}\)
- Với \(t=-3\Rightarrow t^2=9\Rightarrow\left(x-2\right)\left(x+2\right)=9\)
\(\Rightarrow x^2=13\Rightarrow x=-\sqrt{13}\)
176a.
ĐKXĐ: \(0< x\le4\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=a>0\\\sqrt{2-\sqrt{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=4\)
Pt trở thành:
\(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Rightarrow a^2\left(\sqrt{2}-b\right)+b^2\left(\sqrt{2}+a\right)=\sqrt{2}\left(\sqrt{2}-a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)+ab\left(b-a\right)=2\sqrt{2}-\sqrt{2}ab+2\left(a-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab+ab\left(b-a\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\sqrt{2}\left(ab+2\right)+\left(b-a\right)\left(ab+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ab+2=0\\b-a+\sqrt{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4-x}+2=0\left(vn\right)\\a-b=\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
\(\Rightarrow4-2\sqrt{4-x}=2\)
\(\Rightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\) (thỏa mãn)
