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Bài 1:
Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:
\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)
\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc
\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)
Câu 3: ĐK: \(x\ge0\)
Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)
Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)
\(\Rightarrow2x-4\sqrt{x-1}=0\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)
A=(\(3\sqrt{3}-2\sqrt{3}+6\)).\(\sqrt{3}-4\sqrt{3}\)
=\(\sqrt{3}\left(3-2+2\sqrt{3}\right)\).\(\sqrt{3}-4\sqrt{3}\)
=3(\(3-2+2\sqrt{3}\))-4\(\sqrt{3}\)
=3+2\(\sqrt{3}\)
\(\left\{{}\begin{matrix}\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}-4\sqrt{y}=-4\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=1\\\sqrt{x}-2\sqrt{y}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}-2\sqrt{1}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}-2=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\sqrt{x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\) vậy \(y=1;x=0\)
\(\left\{{}\begin{matrix}\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}-3\sqrt{y}=-3\end{matrix}\right.\)
Đặt \(u=\sqrt{x};v=\sqrt{y}\) cho đơn giản (\(u;v\ge0\))
\(\left\{{}\begin{matrix}u-2v=-2\\2u-3v=-3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2u-4v=-4\\2u-3v=-3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-v=-1\\u-2v=-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}v=1\\u-2.1=-2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}v=1\\u=0\end{matrix}\right.\left(TMĐK\right)\)
\(\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
đặt \(\sqrt{x-2}\)=a \(\sqrt{y-3}\)=b
2a+3b=14
\(\left\{{}\begin{matrix}2a+3b=14\\a+b=5\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-b\\2.\left(5-b\right)+3b=14\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=5-b\\10-2b+3b=14\end{matrix}\right.\)\(\left\{{}\begin{matrix}a=5-b\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-4\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=1\\b=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{1}\\\sqrt{y-3}=\sqrt{16}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2=1\\y-3=16\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=3\\y=19\end{matrix}\right.\)
⇒phương trình có 2 no (x,y)=(3, 19)
e cảm ơn