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ĐKXĐ: x > y
Ta có hệ \(\hept{\begin{cases}\sqrt{x+y}+\sqrt{x-y}=4\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y+2\sqrt{\left(x+y\right)\left(x-y\right)}+x-y=16\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x^2-y^2}=16-2x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x^2-y^2}=8-x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}8-x\ge0\\x^2-y^2=\left(8-x\right)^2\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\x^2-y^2=64-16x+x^2\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\-y^2=64-16x\\x^2+y^2=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\\y^2=16x-64\\x^2+y^2-y^2=18-16x+64\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le8\left(1\right)\\y^2=16x-64\left(2\right)\\x^2+16x-82=0\left(3\right)\end{cases}}\)
Giải (3) \(x^2+16x-82=0\)
\(\Leftrightarrow x^2+16x+64=146\)
\(\Leftrightarrow\left(x+8\right)^2=146\)
\(\Leftrightarrow x+8=\pm\sqrt{146}\)
\(\Leftrightarrow x=\pm\sqrt{146}-8\)(Thỏa mãn (1) )
Thay vào (2) tìm được y rồi so sánh ĐKXĐ => KL
@Fabulous Joker cảm ơn ông nhiều lắm
mai tôi phải nộp bài r
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
Áp dụng BĐT AM-GM ta có:
\(VT=\sqrt{x^2+x-5}+\sqrt{-x^2+x+3}\)
\(\le\frac{x^2+x-5+1}{2}+\frac{-x^2+x+3+1}{2}\)
\(=\frac{x^2+x-4}{2}+\frac{-x^2+x+4}{2}=x\)
\(\Rightarrow x\le x^2-3x+2\Leftrightarrow-\left(x-2\right)^2+2\le0\)
Khi \(x=2\pm\sqrt{2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\x\ge-3\\x\ge-4\\x\ge-7\end{matrix}\right.\Leftrightarrow}x\ge-2\)
\(\sqrt{x+2}-\sqrt{x+3}=\sqrt{x+4}-\sqrt{x+7}\)
\(\Leftrightarrow x+2-2\sqrt{\left(x+2\right)\left(x+3\right)}+x+3=x+4-2\sqrt{\left(x+4\right)\left(x+7\right)}+x+7\)
\(\Leftrightarrow-2\sqrt{\left(x+2\right)\left(x+3\right)}+2\sqrt{\left(x+4\right)\left(x+7\right)}=6\)
\(\Leftrightarrow2\left[\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}\right]=6\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}=3\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}+\left(x+2\right)\left(x+3\right)=9\)
\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=-2x^2-16x-8\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=x^2+8x+4\)
Có lẽ làm sai ở đâu đó, mk lười :V
ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+7}=\sqrt{x+3}+\sqrt{x+4}\)
\(\Leftrightarrow2x+9+2\sqrt{x^2+9x+14}=2x+7+2\sqrt{x^2+7x+12}=0\)
\(\Leftrightarrow\sqrt{x^2+9x+14}+1=\sqrt{x^2+7x+12}\)
\(\Leftrightarrow x^2+9x+15+2\sqrt{x^2+9x+14}=x^2+7x+12\)
\(\Leftrightarrow2\sqrt{x^2+9x+14}=-2x-3\) (\(x\le-\frac{3}{2}\))
\(\Leftrightarrow4\left(x^2+9x+14\right)=4x^2+12x+9\)
\(\Leftrightarrow24x=-47\)
\(\Leftrightarrow x=-\frac{47}{24}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(\Leftrightarrow\left|x-2\right|=2-\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2-\sqrt{3}\\x-2=\sqrt{3}-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
PT có tập nghiệm : \(S=\left\{4-\sqrt{3};\sqrt{3}\right\}\)
\(\sqrt{x^2-4x+4}=\sqrt{7-4\sqrt{3}}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(\Leftrightarrow\left|x-2\right|=2-\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2-\sqrt{3}\\-x+2=2-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
Vậy \(x=\sqrt{3}\) hoặc \(x=4-\sqrt{3}\)