giải các phương trình sau ( mình đang cần gấp cảm ơn )
1) x+\(\sqrt{4-x^2}\)= 2+2x.\(\sqrt{4-x^2}\)
2) \(\sqrt{2x^2+11x+19}\)+\(\sqrt{2x^2+5x+1}\)=3.( x+1)
3) \(\sqrt{4x^2+5x+1}\)- 2\(\sqrt{x^2-x+1}\)=9x-3
4) \(\sqrt{2x^2+7x+10}\)+\(\sqrt{2x^2+x+4}\)= 3( x+1)
5) 2x2+5x-1=7.\(\sqrt{x^3-1}\)
6) 2x2+4 = 3\(\sqrt{x^3+1}\)
7) 10\(\sqrt{x^3+1}\)= 3x2+6
1.
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Đặt \(\left\{{}\begin{matrix}x=a\\\sqrt{4-x^2}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a^2+b^2=4\\a+b=2+2ab\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=4\\2ab=a+b-2\end{matrix}\right.\)
Cộng vế với vế:
\(\left(a+b\right)^2=a+b+2\)
\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\Rightarrow\left[{}\begin{matrix}a+b=-1\\a+b=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{4-x^2}=-1\\x+\sqrt{4-x^2}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4-x^2}=-1-x\left(x\le-1\right)\\\sqrt{4-x^2}=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x^2=x^2+2x+1\\4-x^2=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+2x-3=0\\2x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-1-\sqrt{7}}{2}\\x=0\\x=2\end{matrix}\right.\)
3/
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\(\Leftrightarrow\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2+5x+1}=a\ge0\\\sqrt{4x^2-4x+4}=b>0\end{matrix}\right.\) ta được:
\(a-b=a^2-b^2\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)
Th1: \(a=b\Leftrightarrow4x^2+5x+1=4x^2-4x+4\)
\(\Rightarrow9x=3\Rightarrow x=\frac{1}{3}\)
Th2: \(a+b=1\Leftrightarrow\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}=1\)
Mà \(\sqrt{4x^2-4x+4}=\sqrt{\left(2x-1\right)^2+3}\ge\sqrt{3}>1\)
\(\Rightarrow\) Pt vô nghiệm
4.
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+7x+10}=a>0\\\sqrt{2x^2+x+4}=b>0\end{matrix}\right.\) ta được:
\(a+b=\frac{a^2-b^2}{2}\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=2\left(a+b\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\left(loại\right)\\a-b=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+7x+10}-\sqrt{2x^2+x+4}=2\)
\(\Leftrightarrow\sqrt{2x^2+7x+10}=\sqrt{2x^2+x+4}+2\)
\(\Leftrightarrow2x^2+7x+10=2x^2+x+8+4\sqrt{2x^2+x+4}\)
\(\Leftrightarrow3x+1=2\sqrt{2x^2+x+4}\) (\(x\ge-\frac{1}{3}\))
\(\Leftrightarrow\left(3x+1\right)^2=4\left(2x^2+x+4\right)\)
\(\Leftrightarrow x^2+2x-15=0\Rightarrow x=3\)
5.
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\(\Leftrightarrow2x^2+5x-1=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=b\ge0\\\sqrt{x^2+x+1}=a>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=7ab\)
\(\Leftrightarrow2a^2-7ab+3b^2=0\)
\(\Leftrightarrow\left(2a-b\right)\left(a-3b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2a=b\\a=3b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2+x+1}=\sqrt{x-1}\\\sqrt{x^2+x+1}=3\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+x+1\right)=x-1\\x^2+x+1=9x-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2+3x+5=0\left(vn\right)\\x^2-8x+10=0\left(casio\right)\end{matrix}\right.\)
6.
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\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
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)
7.
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\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
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)
