Thôi liên hợp cho nhanh :v :v \(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\)(đk\(-1\le x\le4\) )
<=> \(\sqrt{x+1}-\left(\dfrac{1}{3}x+1\right)+\sqrt{4-x}-\left(2-\dfrac{1}{3}x\right)\)
\(+\sqrt{\left(x+1\right)\left(4-x\right)}-2=0\)
<=> \(\dfrac{1}{3}\left[\dfrac{9\left(x+1\right)-\left(x+3\right)^2}{\sqrt{x+1}+\left(x+3\right)}\right]+\dfrac{1}{3}\left[\dfrac{9\left(4-x\right)-\left(6-x\right)^2}{\sqrt{4-x}+\left(6-x\right)}\right]\)
+ \(\dfrac{\left(x+1\right)\left(4-x\right)-4}{\sqrt{\left(x+1\right)\left(4-x\right)}+2}=0\)
<=> \(\dfrac{3x-x^2}{3\left[\sqrt{x+1}+\left(x+3\right)\right]}+\dfrac{3x-x^2}{3\left[\sqrt{4-x}+\left(6-x\right)\right]}\)
\(+\dfrac{3x-x^2}{\sqrt{\left(x+1\right)\left(4-x\right)}+2}=0\)
<=> \(\left(3x-x^2\right)\left(\dfrac{1}{3\left[\sqrt{x+1}+\left(x+3\right)\right]}+\dfrac{1}{3\left[\sqrt{4-x}+\left(6-x\right)\right]}+\dfrac{1}{\sqrt{\left(x+1\right)\left(4-x\right)}+2}\right)=0\)
Dễ thấy \(\dfrac{1}{3\sqrt{x+1}+3\left(x+3\right)}+\dfrac{1}{3\sqrt{4-x}+3\left(6-x\right)}+\dfrac{1}{\sqrt{\left(x+1\right)\left(4-x\right)}+2>0\forall-1\le x\le4}\)
=> \(x\left(3-x\right)=0\)
<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) ( TM)
P/s: Hiểu ko???????????
Dù câu hỏi đăng lâu r ( 4 ngày trước ) cơ mà thấy bạn nói là đặt ẩn phụ không được .... nên mik sẽ dùng phương pháp đặt ẩn phụ để giải bài này
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- ĐK \(\left\{{}\begin{matrix}x+1\ge0,4-x\ge0\\\left(x-1\right)\left(4-x\right)\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le4\\\left(x-1\right)\left(4-x\right)\ge0\end{matrix}\right.\)
- Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{4-x}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow\left\{{}\begin{matrix}x+1=a^2\\4-x=b^2\end{matrix}\right.\) \(\Rightarrow a^2+b^2=5\)
Theo bài ra ta có : a + b+ ab=5
\(\Rightarrow\left\{{}\begin{matrix}a+b=5-ab\\a^2+b^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2=\left(5-ab\right)^2\\\left(a+b\right)^2=5+2ab\end{matrix}\right.\)
\(\Leftrightarrow25-10ab+a^2b^2=5+2ab\)
\(\Leftrightarrow a^2b^2-12ab+20=0\)
\(\Leftrightarrow\left(ab-10\right)\left(ab-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}ab=10\\ab=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+ab=a+b+10=5\\\sqrt{\left(x+1\right)\left(4-x\right)}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=-5\left(loại\right)\\-x^2+3x+4=4\end{matrix}\right.\) => x2-3x=0
<=> x(x-3)=0 <=> x=3 hoặc x=0
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{4-x}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\) thì ta có:
\(a+b+ab=5\)
\(\Leftrightarrow a+1+b\left(a+1\right)=6\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)=6\)
Đặt như của mình cái ra luôn mà :v
\(PT\Leftrightarrow a+\dfrac{a^2-5}{2}=5\)
