Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\\b=\sqrt{9-2x}\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow ab=\sqrt{-2x^2+15x-27};a^2+b^2=-x+6\) \(PT\Leftrightarrow a+b=2ab+a^2+b^2+2\\ \Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)^2+\left(a+b\right)-2\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)\left(a+b+1\right)-2\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b-2\right)\left(a+b+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=2\left(n\right)\\a+b=-1\left(l\right)\end{matrix}\right.\) \(\Leftrightarrow a+b=2\\ \Leftrightarrow\sqrt{x-3}=2-\sqrt{9-2x}\left(3\le x\le\dfrac{9}{2}\right)\\ \Leftrightarrow x-3=11-2x-4\sqrt{9-2x}\\ \Leftrightarrow x+14=4\sqrt{9-2x}\\ \Leftrightarrow x^2+28x+196=144-32x\\ \Leftrightarrow x^2+60x+52=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-30+4\sqrt{53}\left(l\right)\\x=-30-4\sqrt{53}\left(l\right)\end{matrix}\right.\) Vậy PT vô nghiệm Câu trả lời: Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\\b=\sqrt{9-2x}\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow ab=\sqrt{-2x^2+15x-27};a^2+b^2=-x+6\) \(PT\Leftrightarrow a+b=2ab+a^2+b^2+2\\ \Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)^2+\left(a+b\right)-2\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)\left(a+b+1\right)-2\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b-2\right)\left(a+b+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=2\left(n\right)\\a+b=-1\left(l\right)\end{matrix}\right.\) \(\Leftrightarrow a+b=2\\ \Leftrightarrow\sqrt{x-3}=2-\sqrt{9-2x}\left(3\le x\le\dfrac{9}{2}\right)\\ \Leftrightarrow x-3=11-2x-4\sqrt{9-2x}\\ \Leftrightarrow x+14=4\sqrt{9-2x}\\ \Leftrightarrow x^2+28x+196=144-32x\\ \Leftrightarrow x^2+60x+52=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-30+4\sqrt{53}\left(l\right)\\x=-30-4\sqrt{53}\left(l\right)\end{matrix}\right.\) Vậy PT vô nghiệm

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Vũ Thu Trang

20 tháng 11 2021

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Nguyễn Hoàng Minh

20 tháng 11 2021

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\\b=\sqrt{9-2x}\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow ab=\sqrt{-2x^2+15x-27};a^2+b^2=-x+6\)

\(PT\Leftrightarrow a+b=2ab+a^2+b^2+2\\ \Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)^2+\left(a+b\right)-2\left(a+b\right)-2=0\\ \Leftrightarrow\left(a+b\right)\left(a+b+1\right)-2\left(a+b+1\right)=0\\ \Leftrightarrow\left(a+b-2\right)\left(a+b+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b=2\left(n\right)\\a+b=-1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow a+b=2\\ \Leftrightarrow\sqrt{x-3}=2-\sqrt{9-2x}\left(3\le x\le\dfrac{9}{2}\right)\\ \Leftrightarrow x-3=11-2x-4\sqrt{9-2x}\\ \Leftrightarrow x+14=4\sqrt{9-2x}\\ \Leftrightarrow x^2+28x+196=144-32x\\ \Leftrightarrow x^2+60x+52=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-30+4\sqrt{53}\left(l\right)\\x=-30-4\sqrt{53}\left(l\right)\end{matrix}\right.\)

Vậy PT vô nghiệm

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