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ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)
\(\Leftrightarrow2\sqrt{2x-5}=10\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow2x-5=25\)
\(\Leftrightarrow x=15\)
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\(\dfrac{3\sqrt{x}}{2}-\dfrac{2\sqrt{x}-7}{3}=\sqrt{x}-1\)
\(\Leftrightarrow9\sqrt{x}-15-4\sqrt{x}+14=6\sqrt{x}-6\left(x\ge0\right)\)
\(\Leftrightarrow5\sqrt{x}-1=6\sqrt{x}-6\)
\(\Leftrightarrow x=25\left(TM\right)\)
KL.....
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Nguyễn Huy TúAkai HarumaLightning FarronMysterious PersonDƯƠNG PHAN KHÁNH DƯƠNG
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Đặt \(2x-5=t^2\)ta có \(x=\frac{t^2+5}{2}\)thay giá trị của x vào phương trình đã cho được:
\(\sqrt{\frac{t^2+5}{2}-2+t}+\sqrt{\frac{t^2+5}{2}+2+3t}=7\sqrt{2}\)
hay \(\sqrt{t^2+5-2+2t}+\sqrt{t^2+5+4+6t}=14\)
\(\sqrt{t^2+2t+1}+\sqrt{t^2+6t+9}=14\)
\(\sqrt{\left(t+1\right)^2}+\sqrt{\left(t+3\right)^2}=14\)
\(t+1+t+3=14\)
\(2t+4=14\)
2t=10
t=5
Từ đó \(x=\frac{25+5}{2}=15\)
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Đặt \(\sqrt[3]{7-x}=a;\sqrt[3]{5-x}=b\) ( a + b \(\ne\) 0)
=> a3 + b3 = 12 - 2x = 2(6 - x) ; a3 - b3 = 2
PT <=> \(\frac{a-b}{a+b}=\frac{a^3+b^3}{2}\) <=> (a3 + b3)(a+ b) = 2(a - b)
Thế 2 = a3 - b3 ta được:
(a3 + b3)(a+ b) = (a3 - b3)(a - b)
<=> a4 + a3b + ab3 + b4 = a4 - a3b - ab3 + b4
<=> a3b + ab3 = - a3b - ab3
<=> 2(a3b + ab3) = 0 <=> ab.(a2+ b2) = 0 <=> ab = 0 hoặc a2 + b2 = 0
+) ab = 0 => a = 0 hoặc b = 0
Nếu a = 0 thì b3 = - 2 => \(b=-\sqrt[3]{2}\)
Nếu b = 0 thì a3 = 2 => \(a=\sqrt[3]{2}\)
+) a2 + b2 = 0 => a = b = 0 => Loại (vì a + b khác 0)
Vậy a = 0 hoặc b = 0
a = 0 => x = 7
b = 0 => x = 5
Vậy...........
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<=>\(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}-3y=3\\\sqrt{5}x-2y=7\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)
KL: vậy hpt có ngiệm là \(\left\{{}\begin{matrix}x=\sqrt{5}\\y=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
ĐKXĐ: \(x\ne\pm y\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\2u+3v=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u+3v=5\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=2-u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=1\\\dfrac{1}{x-y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+7=x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-5x+6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
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ĐKXĐ : \(\frac{5}{3}\le x\le12\)
\(2\sqrt{3x-5}-3\sqrt{12-x}+x^2+x-7=0\)
\(\Leftrightarrow\left(2\sqrt{3x-5}-4\right)+\left(9-3\sqrt{12-x}\right)+x^2+x-12=0\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{2\sqrt{3x-5}+4}+\frac{9\left(x-3\right)}{9+3\sqrt{12-x}}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{2\sqrt{3x-5}+4}+\frac{9}{9+3\sqrt{12-x}}+x+4\right)=0\)
\(\Leftrightarrow x=3\)( vì vế trong ngoặc thứ 2 > 0 \(\forall\)\(\frac{5}{3}\le x\le12\))
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