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a)
Pt\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=\left(x-3\right)^2\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4=x^2-6x+9\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+13=0\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{9+\sqrt{29}}{2}\\x_2=\dfrac{9-\sqrt{29}}{2}\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{9+\sqrt{29}}{2}\)
Vậy \(x=\dfrac{9+\sqrt{29}}{2}\) là nghiệm của phương trình.
b) Pt \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+3=\left(2x-1\right)^2\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-2=0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{7}}{3}\\x_2=\dfrac{1-\sqrt{7}}{3}\end{matrix}\right.\\x\ge\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{7}}{3}\)
Vậy phương trình có duy nhất nghiệm là: \(x=\dfrac{1+\sqrt{7}}{3}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))
\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\le12\)
\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)
\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)
Bài 1:
ĐKXĐ: \(2-3x>0\Rightarrow x< \frac{2}{3}\)
\(\Leftrightarrow3x-m+5+2-3x=2x+2m-1\)
\(\Leftrightarrow2x=8-3m\Rightarrow x=\frac{8-3m}{2}\)
Để pt đã cho có nghiệm
\(\Rightarrow\frac{8-3m}{2}< \frac{2}{3}\Leftrightarrow24-9m< 4\Rightarrow m>\frac{20}{9}\)
Bài 2:
\(\Leftrightarrow\left(x-2\right)^4+4\left(x^2+2x-1\right)^4-5\left(x-2\right)^2\left(x^2+2x-1\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\left(x-2\right)^2=a\ge0\\\left(x^2+2x-1\right)^2=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2-5ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=\left(x^2+2x-1\right)^2\\\left(x-2\right)^2=4\left(x^2+2x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x^2+2x-1+x-2\right)\left(x^2+2x-1-x+2\right)=0\\\left(2x^2+4x-2+x-2\right)\left(2x^2+4x-2-x+2\right)=0\end{matrix}\right.\)
Bạn tự giải nốt, dạng cơ bản
ĐKXĐ: \(-2\le x\le8\)
\(\Leftrightarrow\sqrt{x+2}-3+1-\sqrt{8-x}=3x^3-21x^2+2x-14\)
\(\Leftrightarrow\frac{x-7}{\sqrt{x+2}+3}+\frac{x-7}{1+\sqrt{8-x}}=\left(x-7\right)\left(3x^2+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\frac{1}{\sqrt{x+2}+3}+\frac{1}{1+\sqrt{8-x}}=3x^2+2\left(1\right)\end{matrix}\right.\)
Xét (1), do \(\left\{{}\begin{matrix}\sqrt{x+2}\ge0\\\sqrt{8-x}\ge0\end{matrix}\right.\) \(\Rightarrow VT< \frac{1}{3}+1< 2\)
\(VP=3x^2+2\ge2>VT\)
\(\Rightarrow\) (1) vô nghiệm
Vậy pt có nghiệm duy nhất \(x=7\)