Câu 2b đề là tìm x chứ nhỉ???

b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2

Vậy x = 2

Có giải mà!

a) Những oxit: CaO; SO₂; SO₃; CO₂; CO

b) Những oxit: CO ( H₂SO₄ loãng) ; Fe₂O₃ ; CuO; Al₂O_3; CaO

c) Những oxit: SO₂; SO₃; CO; CO₂

\(x=\sqrt{9+2.3.2\sqrt{5}+\left(2\sqrt{5}\right)^2}+\sqrt{9-2.3.2\sqrt{5}+\left(2\sqrt{5}\right)^2}\)

= \(\sqrt{\left(3+2\sqrt{5}\right)^2}+\sqrt{\left(3-2\sqrt{5}\right)2}\) = \(3+2\sqrt{5}+3-2\sqrt{5}\) = 6

=> x = 6

Ta có: \(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

Đặt x + y = a ; xy = b

=> (1) trở thành: \(\left\{{}\begin{matrix}a+b=7\left(2\right)\\a^2-b=13\left(3\right)\end{matrix}\right.\)

Cộng (2) với (3) ta được: a² + a = 20

<=> a² + a - 20 = 0

<=> (a - 4)(a + 5) = 0

<=> \(\left[{}\begin{matrix}a=4\\a=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=-5\\b=12\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-5\\xy=12\end{matrix}\right.\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4-y\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-5-y\\xy=12\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}\left(4-y\right)y=3\\-\left(5+y\right)y=12\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}y^2-4y+3=0\\y^2+5y+12=0\end{matrix}\right.\)

<=> y² - 4y + 3 = 0 ( vì y² + 5y + 12 = 0 vô nghiệm)

<=> ( y - 1)(y - 3) = 0

<=> \(\left[{}\begin{matrix}y=1\\y=3\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\end{matrix}\right.\)

Vậy ( x;y) = ( 3;1), (1;3)