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5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)
f/
ĐKXĐ: ...
Đặt \(\sqrt{2-x}+\sqrt{x+2}=a>0\)
\(\Rightarrow a^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}\)
Phương trình trở thành:
\(a+\frac{a^2-4}{2}=2\)
\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}=0\)
\(\Rightarrow4-x^2=0\Rightarrow x=\pm2\)
e/ ĐKXĐ: ...
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\)
\(\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\)
Pt trở thành:
\(a+\frac{a^2-5}{2}=5\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=2\)
\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Câu 6:
ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)-2\sqrt{x-1}+1}-\sqrt{x-1}=1$
$\Leftrightarrow \sqrt{(\sqrt{x-1}-1)^2}=\sqrt{x-1}+1$
$\Leftrightarrow |\sqrt{x-1}-1|=\sqrt{x-1}+1$
Nếu $\sqrt{x-1}-1\geq 0$ thì PT trở thành:
$\sqrt{x-1}-1=\sqrt{x-1}+1\Leftrightarrow 2=0$ (vô lý)
Nếu $\sqrt{x-1}-1< 0$ (tương đương với $1\leq x< 2$ thì PT trở thành:
$1-\sqrt{x-1}=\sqrt{x-1}+1$
$\Leftrightarrow \sqrt{x-1}=0\Rightarrow x=1$ (thỏa mãn)
Vậy PT có nghiệm $x=1$
Câu 5:
ĐK: $x\geq 1$
PT $\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1$
$\Leftrightarrow \sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1$
$\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=|\sqrt{x-1}-2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}-2+3-\sqrt{x-1}|=1$
Dấu "=" xảy ra khi $(\sqrt{x-1}-2)(3-\sqrt{x-1})\geq 0$
$\Leftrightarrow 3\geq \sqrt{x-1}\geq 2$
$\Leftrightarrow 10\geq x\geq 5$. Kết hợp ĐKXĐ ta thấy những giá trị $x$ thỏa mãn $10\geq x\geq 5$ là nghiệm của pt.
5.
\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)
Câu 6 bạn coi lại đề
4.
ĐKXĐ: ...
Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)
\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)
\(\Rightarrow2x^2-ax-a^2=0\)
\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)
5.
ĐKXĐ: ...
\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)
\(\Leftrightarrow x=5\)
6.
ĐKXĐ: \(-4\le x\le4\)
\(\Leftrightarrow\frac{\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\frac{x\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4-x}+2=2\sqrt{x+4}+4\)
\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)
\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)
\(\Rightarrow x=\frac{96}{25}\)
1.
Bạn coi lại đề
2.
ĐKXĐ: \(1\le x\le2\)
Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:
\(\left(\sqrt{x+2}+\sqrt{x-1}\right)\left(\sqrt{x+2}-\sqrt{x-1}\right)\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+2-\sqrt{x+2}+1-\sqrt{x-1}=0\)
\(\Leftrightarrow3\sqrt{2-x}+\frac{2-x}{2+\sqrt{x+2}}+\frac{2-x}{1+\sqrt{x-1}}=0\)
\(\Leftrightarrow\sqrt{2-x}\left(3+\frac{\sqrt{2-x}}{2+\sqrt{x+2}}+\frac{\sqrt{2-x}}{1+\sqrt{x-1}}\right)=0\)
\(\Leftrightarrow\sqrt{2-x}=0\Rightarrow x=2\)