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ĐK: \(\left\{{}\begin{matrix}x\ne-y\\y\ge\dfrac{3}{2}\end{matrix}\right.\).
\(\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}=1\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-1=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-y+3}{x+y}-\dfrac{x+y}{x+y}=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y+3-x-y=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y+3=0\\2x-\sqrt{2y-3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\left(2y-3\right)=0\\2x-\sqrt{2y-3}=0\end{matrix}\right..\)
Đặt a = x, b = \(\sqrt{2y-3}\).
Hệ phương trình trở thành: \(\left\{{}\begin{matrix}a-b^2=0\\2a-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\2b^2-b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\b\left(2b-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}a=0\\a=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}b=0\\b=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y-3=\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\2y=\dfrac{13}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=\dfrac{13}{8}\end{matrix}\right.\end{matrix}\right..\)
Vậy hệ phương trình có nghiệm (x;y) \(\in\) \(\left\{\left(0;\dfrac{3}{2}\right),\left(\dfrac{1}{4};\dfrac{13}{8}\right)\right\}\).
\(x^4+3x^2=0\)
Có \(x^4\ge0;\forall x\); \(3x^2\ge0;\forall x\)
=> VT\(\ge0;\forall x\)
Dấu = xảy ra <=> x=0
Ý C
Câu 3:
a) Ta có: \(x^2-1=3\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\sqrt{16x}-2\sqrt{36x}+\sqrt{9x}=2\)
\(\Leftrightarrow4\sqrt{x}-12\sqrt{x}+3\sqrt{x}=2\)
\(\Leftrightarrow-5\sqrt{x}=2\)(Vô lý)
\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)
Câu 4:
Thay x=2 và y=-1 vào hệ, ta được:
\(\left\{{}\begin{matrix}2a-b=4\\2b+2=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2\\a=1\end{matrix}\right.\)
Câu 4:
\(a,\tan B=\dfrac{AC}{AB}=\dfrac{12}{5}\approx\tan67^0\\ \Rightarrow\widehat{B}\approx67^0\\ b,\text{Áp dụng PTG: }BC=\sqrt{AC^2+AB^2}=13\left(cm\right)\\ \text{Áp dụng HTL: }\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{25}{13}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{144}{13}\left(cm\right)\\AH=\sqrt{BH\cdot CH}=\dfrac{60}{13}\left(cm\right)\end{matrix}\right.\)