Chương I - Căn bậc hai. Căn bậc ba

HC

Giải các phương trình:

a) \(x^2+6x-3=4x\sqrt{2x-1}\)

b) \(3x^2+4x-3=4x\sqrt{4x-3}\)

c) \(3x^2+2x+7=3\left(x+1\right)\sqrt{x^2+3}\)

d) \(6x^2+4x+8=5\left(x+1\right)\sqrt{2x^2+3}\)

e)\(6x^2-x=21+\left(x-3\right)\sqrt{x^2+x-6}\)

NL
23 tháng 6 2019 lúc 23:18

a/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)

\(\Leftrightarrow\frac{4x\left(x-1\right)^2}{x+\sqrt{2x-1}}-3\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{4x}{x+\sqrt{2x-1}}=3\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)

\(\Leftrightarrow x=3\sqrt{2x-1}\)

\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)

Vậy pt có 3 nghiệm....

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NL
23 tháng 6 2019 lúc 23:22

b/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow\frac{4x\left(x^2-4x+3\right)}{x+\sqrt{4x-3}}-\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\Rightarrow x=...\\\frac{4x}{x+\sqrt{4x-3}}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)

\(\Leftrightarrow3x=\sqrt{4x-3}\)

\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)

Vậy...

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NL
23 tháng 6 2019 lúc 23:27

c/

\(\Leftrightarrow x^2+2x+1+2\left(x^2+3\right)=3\left(x+1\right)\sqrt{x^2+3}\)

\(\Leftrightarrow\left(x+1\right)^2+2\left(x^2+3\right)-3\left(x+1\right)\sqrt{x^2+3}=0\)

Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{x^2+3}=b\end{matrix}\right.\)

\(\Rightarrow a^2+2b^2-3ab=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\2\sqrt{x^2+3}=x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\\4x^2+12=x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\3x^2-2x+11=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=1\)

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NL
23 tháng 6 2019 lúc 23:34

d/

\(\Leftrightarrow2\left(2x^2+3\right)+2\left(x^2+2x+1\right)-5\left(x+1\right)\sqrt{2x^2+3}=0\)

\(\Leftrightarrow2\left(2x^2+3\right)+2\left(x+1\right)^2-5\left(x+1\right)\sqrt{2x^2+3}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)

\(\Rightarrow2a^2+2b^2-5ab=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{2x^2+3}=x+1\\\sqrt{2x^2+3}=2\left(x+1\right)\end{matrix}\right.\) (\(x\ge-1\))

\(\Leftrightarrow\left[{}\begin{matrix}8x^2+12=x^2+2x+1\\2x^2+3=4x^2+8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x^2-2x+11=0\left(vn\right)\\2x^2+8x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-4+\sqrt{14}}{2}\\x=\frac{-4-\sqrt{14}}{2}\left(l\right)\end{matrix}\right.\)

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NL
23 tháng 6 2019 lúc 23:44

e/ ĐKXĐ: \(x^2+x-6\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\)

\(\Leftrightarrow5\left(x^2+x-6\right)-\left(x-3\right)\sqrt{x^2+x-6}+\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow5\left(x^2+x-6\right)-\left(x-3\right)\sqrt{x^2+x-6}+\left(x-3\right)^2=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x-6}=a\\x-3=b\end{matrix}\right.\)

\(\Rightarrow5a^2-ab+b^2=0\)

\(\Leftrightarrow\frac{19}{4}a^2+\left(\frac{a}{2}-b\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-6=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow ptvn\)

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