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Bài 2:
ĐKXĐ: $6\geq x\geq \frac{-1}{3}$
PT $\Leftrightarrow (\sqrt{3x+1}-4)+(1-\sqrt{6-x})+(3x^2-14x-5)=0$
$\Leftrightarrow \frac{3(x-5)}{\sqrt{3x+1}+4}+\frac{x-5}{\sqrt{6-x}+1}+(3x+1)(x-5)=0$
$\Leftrightarrow (x-5)\left[\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+(3x+1)\right]=0$
Với $x$ thuộc đkxđ, dễ thấy biểu thức trong ngoặc vuông $>0$
$\Rightarrow x-5=0$
$\Leftrightarrow x=5$
Bài 3:
PT $3x=\sqrt{x^2+12}-\sqrt{x^2+5}+5>0$
$\Rightarrow x>0$
Lại có:
PT $\Leftrightarrow \sqrt{x^2+12}-4=3(x-2)+(\sqrt{x^2+5}-3)$
$\Leftrightarrow \frac{x^2-4}{\sqrt{x^2+12}+4}=3(x-2)+\frac{x^2-4}{\sqrt{x^2+5}+3}$
$\Leftrightarrow (x-2)\left[\frac{x+2}{\sqrt{x^2+12}+4}-3-\frac{x+2}{\sqrt{x^2+5}+3}\right]=0$
Với $x>0$, dễ thấy:
$\frac{x+2}{\sqrt{x^2+5}+3}+3>\frac{x+2}{\sqrt{x^2+12}+4}$ nên biểu thức trong ngoặc vuông âm.
Do đó $x-2=0\Leftrightarrow x=2$ (tm)
a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)
\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)
\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)
pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)
suy ra tìm đc x
a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)
đặt\(x^2+x+1=t\left(t>0\right)\)
\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)
bình phương 2 vế pt trở thành:
\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)
\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m
vậy pt vô nghiệm
a/ ĐKXĐ: ...
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)
\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)
\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))
\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)
\(\Leftrightarrow11a^2+6a-25=0\)
Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó
b/
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)
\(\Leftrightarrow\sqrt{a^2+3a}=2\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
a/ \(\Leftrightarrow x^2+5x-2-2\sqrt[3]{x^2+5x-2}+4=0\)
Đặt \(\sqrt[3]{x^2+5x-2}=a\)
\(a^3-2a+4=0\)
\(\Leftrightarrow\left(a+2\right)\left(a^2-2a+2\right)=0\Rightarrow a=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\Rightarrow x^2+5x+6=0\Rightarrow...\)
b/ ĐKXĐ:...
\(\Leftrightarrow-3\left(-x^2+4x+10\right)-5\sqrt{-x^2+4x+10}+42=0\)
Đặt \(\sqrt{-x^2+4x+10}=a\ge0\)
\(-3a^2-5a+42=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{14}{3}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+4x+10}=3\Rightarrow x^2-4x-1=0\Rightarrow...\)
c/ ĐKXĐ: ...
\(\Leftrightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)
Đặt \(\sqrt{x^2+3x}=a\ge0\)
\(a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+3x}=2\Rightarrow x^2+3x-4=0\)
d/ ĐKXĐ: \(-1\le x\le2\)
\(\Leftrightarrow\sqrt{3-x+x^2}=1+\sqrt{2+x-x^2}\)
\(\Leftrightarrow3-x+x^2=3+x-x^2+2\sqrt{2+x-x^2}\)
\(\Leftrightarrow2+x-x^2+\sqrt{2+x-x^2}-2=0\)
Đặt \(\sqrt{2+x-x^2}=a\ge0\)
\(a^2+a-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2+x-x^2}=1\Leftrightarrow x^2-x-1=0\)
e/ \(\Leftrightarrow\sqrt{x^2-3x+3}-1+\sqrt{x^2-3x+6}-2=0\)
\(\Leftrightarrow\frac{x^2-3x+2}{\sqrt{x^2-3x+3}+1}+\frac{x^2-3x+2}{\sqrt{x^2-3x+6}+2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{\sqrt{x^2-3x+3}+1}+\frac{1}{\sqrt{x^2-3x+6}+2}\right)=0\)
\(\Leftrightarrow x^2-3x+2=0\)
a/ ĐKXĐ: \(x^2+3x+2\ge0\)
\(\Leftrightarrow3-2\sqrt{x^2+3x+2}=1-2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\sqrt{x^2+3x+2}=\sqrt{x^2-x+1}+1\)
\(\Leftrightarrow x^2+3x+2=x^2-x+1+1+2\sqrt{x^2-x+1}\)
\(\Leftrightarrow2x=\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2-x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x^2+x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{-1+\sqrt{13}}{6}\\x=\frac{-1-\sqrt{13}}{6}\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(3x^2-7x+2\ge0\)
\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\) (1)
\(\Rightarrow3x^2-5x+7=9+3x^2-7x+2-6\sqrt{3x^2-7x+2}\)
\(\Rightarrow2-x=3\sqrt{3x^2-7x+2}\) (\(x\le2\))
\(\Rightarrow\left(2-x\right)^2=9\left(3x^2-7x+2\right)\)
\(\Rightarrow x^2-4x+4=27x^2-63x+18\)
\(\Rightarrow26x^2-59x+14=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)
Do bước biến đổi thứ 2 ko phải phép tương đương nên cần thay 2 nghiệm vào (1) để kiểm tra lại, bạn tự thay nhé
1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
a/ ĐKXĐ: ...
\(\Leftrightarrow\left(x^2-6x\right)\left(\sqrt{17-x^2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x=0\\\sqrt{17-x^2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2=16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\left(l\right)\\x=4\\x=-4\end{matrix}\right.\)
b/ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\left(l\right)\\x=-3\end{matrix}\right.\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\le1\end{matrix}\right.\) \(\Rightarrow x=1\)
Thay \(x=1\) vào pt thấy ko thỏa mãn
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\left(l\right)\\x=2\end{matrix}\right.\)
a, ĐK: \(x\ge\dfrac{1}{5}\)
\(pt\Leftrightarrow\sqrt{5x^2+x+3}+5x-1-2\sqrt{5x-1}+1+x^2+2x+1=-2\)
\(\Leftrightarrow\sqrt{5x^2+x+3}+\left(\sqrt{5x-1}-1\right)^2+\left(x+1\right)^2=-2\)
\(\Rightarrow\) Phương trình vô nghiệm