Giải phương trình \(5\sqrt{x^3+1}=2\left(x^2+2\right)\)
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\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x^2+7x+10}+1\right)=3\)
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{\left(x+5\right)\left(x+2\right)}+1\right)=3\)
Đặt \(\hept{\begin{cases}\sqrt{x+5}=a\left(a\ge0\right)\\\sqrt{x+2}=b\left(b\ge0\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1\right)=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1-a-b\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\end{cases}}\)
Với a = b thì
\(\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow0x=3\left(l\right)\)
Với a = 1 thì
\(\sqrt{x+5}=1\Leftrightarrow x=-4\left(l\right)\)
Với b = 1 thì
\(\sqrt{x+2}=1\Leftrightarrow x=-1\)
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\(\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(1+\sqrt{x^2-x-2}\right)=3\left(DKXD:x\ge2\right)\)\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(\sqrt{x+1}+\sqrt{x-2}\right)\left(1+\sqrt{x\left(x-2\right)+\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)\(\Leftrightarrow\left\{\left(x+1\right)-\left(x-2\right)\right\}\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)
\(\Leftrightarrow3\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)
\(\Leftrightarrow\sqrt{x+1}-\sqrt{\left(x+1\right)\left(x-2\right)}+\sqrt{x-2}-1=0\)
\(\Leftrightarrow-\left(\sqrt{x+1}-1\right)\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=1\\\sqrt{x-2}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\left(loai\right)\\x=3\left(nhan\right)\end{cases}}}\)
Vậy...
Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{x-2}=b\end{cases}}\left(a,b\ge0\right)\) thì ta có
\(\hept{\begin{cases}a^2-b^2=3\left(1\right)\\\left(a-b\right)\left(1+ab\right)=3\left(2\right)\end{cases}}\)
Lấy (1) - (2) vế theo vế ta được
\(a^2-b^2-\left(a-b\right)\left(1+ab\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1-ab\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(1-a\right)\left(b-1\right)=0\)
Với a = b
\(\Leftrightarrow\sqrt{x+1}=\sqrt{x-2}\)
\(\Leftrightarrow x+1=x-2\Leftrightarrow0x=3\left(l\right)\)
Với a = 1
\(\Leftrightarrow\sqrt{x+1}=1\Leftrightarrow x=0\left(l\right)\)
Với b = 1
\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x=3\)
Vậy PT có nghiệm là x = 3
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Đặt x^2+3x=a
=>\(a+2=3\sqrt{a}\)
=>a-3 căn a+2=0
=>(căn a-1)(căn a-2)=0
=>a=1 hoặc a=4
=>x^2+3x=1 hoặc x^2+3x=4
=>(x+4)(x-1)=0 và x^2+3x-1=0
=>\(x\in\left\{1;-4;\dfrac{-3+\sqrt{13}}{2};\dfrac{-3-\sqrt{13}}{2}\right\}\)
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Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
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TUY BẠN CHO ĐỀ HƠI SAI SAI NHƯNG MIK VẪN GIẢI/// ĐÁP ÁN NÈ:
x = 3 !!!!! nếu thiếu thông cảm dùm mik nha
ĐKXĐ: \(x\ge-1\)
\(5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)
Pt trở thành:
\(5ab=2\left(a^2+b^2\right)\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}=\sqrt{x+1}\\\sqrt{x^2-x+1}=2\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)=x+1\\x^2-x+1=4\left(x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)