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ok nha bn

ĐKXĐ: y \(\ne\)0

\(\hept{\begin{cases}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{cases}}\)<=> \(\hept{\begin{cases}\left(x+\frac{1}{y}\right)^2-\frac{2x}{y}+\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{cases}}\)<=> \(\hept{\begin{cases}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{cases}}\)

<=> ^{\(\hept{\begin{cases}\left(x+\frac{1}{y}\right)^2+\left(x+\frac{1}{y}\right)=6\left(1\right)\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{cases}}\)}

Giải pt (1) : Đặt a = \(x+\frac{1}{y}\)

Khi đó ta có pt: a² + a = 6

<=> a² + a - 6 = 0 <=> (a - 2)(a + 3) = 0 <=> \(\orbr{\begin{cases}a=2\\a=-3\end{cases}}\)

* Với a = 2, ta có \(x+\frac{1}{y}\) = 2 => \(\frac{x}{y}=3-2=1\)<=> x = y

Thay x = y vào pt:  \(x+\frac{1}{y}\) = 2  ta dc:

y + \(\frac{1}{y}=2\) <=> y² + 1 = 2y <=> y² - 2y + 1 = 0 <=> (y - 1)² = 0 <=> y = 1 (tmđk) => x = 1

* Với a = -3, ta có \(x+\frac{1}{y}\) = -3 => \(\frac{x}{y}=3+3=6\)<=> x = 6y

Thay x = 6y vào pt:  \(x+\frac{1}{y}=-3\)ta dc:

\(6y+\frac{1}{y}=-3\) <=> 6y² + 1 = -3y <=> 6y² + 3y + 1 = 0 (\(\Delta=-15\)<0 ) (VN)

Vậy nghiệm của hpt là: (1;1)
P/S: xem lại nhé, t làm hơi ẩu

=\(4x-4\sqrt{x}+1+\)3

=\(\left(2\sqrt{x}-1\right)^{^2}\)_{+3\(\ge\)3 với mọi x\(\inℝ\)}

Dấu bằng xảy ra<=>\(2\sqrt{x}-1=0\)<=>\(2\sqrt{x}=1\)

<=> \(\sqrt{x}=\frac{1}{2}\)<=>\(x=\frac{1}{4}\)

Vây minA=3 tại \(x=\frac{1}{4}\)

A, thôi chết nhầm đề rồi

\(\frac{x^2}{\left(x+2\right)^2}+3=3x^2-6x\left(đkxđ:x\ne-2\right)\)

\(< =>\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

\(< =>x^2=\left(3x^2-6x-3\right)\left(x^2+4x+4\right)\)

\(< =>x^2=3x^4+12x^3+12x^2-6x^3-24x^2-14x-3x^2-12x-12\)

\(< =>3x^4+6x^3-16x^2-26x-12=0\)

Đến đây dễ rồi ha !

\(\frac{x^2}{\left(x+2\right)^2}+3=3x^2-6xĐK:x\ne-2\)

\(\Leftrightarrow\frac{x^2}{\left(x+2\right)^2}+\frac{3\left(x+2\right)^2}{\left(x+2\right)^2}=\frac{\left(3x^2-6\right)\left(x+2\right)^2}{\left(x+2\right)^2}\)

Khử mẫu và rút gọn ta đc : \(-3x^4-12x^3-2x^2+36x+36=0\)

Mời nhân tài chứ e chịu r 

B=\(\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{10\sqrt{x}-2x+2\sqrt{x}+3\sqrt{x}-3-x-4\sqrt{x}-\sqrt{x}-4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{-3x+3\sqrt{x}+7\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)

Vậy...

\(B=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x-3}}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)( \(x\ge0;x\ne1\)

=>\(B=\frac{10\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

=> \(B=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

=> \(B=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)

=> \(B=\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)( zì \(x\ge0,x\ne1\)