DK:\(y\ne0\)

PT (1) :\(3x^2+2y^2-4xy=11-\dfrac{1}{y}\left(2x+\dfrac{1}{y}\right)\)

\(\Leftrightarrow\left(x^2+\dfrac{2x}{y}+\dfrac{1}{y^2}\right)+2\left(x^2-2xy+y^2\right)=11\)

\(\Leftrightarrow\left(x+\dfrac{1}{y}\right)^2+2\left(x-y\right)^2=11\)

PT (2): \(2x+\dfrac{1}{y}-y=4\)

\(\Leftrightarrow\left(x+\dfrac{1}{y}\right)+\left(x-y\right)=4\)

Đặt \(a=x+\dfrac{1}{y};b=x-y\)

Hệ pt tt: \(\left\{{}\begin{matrix}a^2+2b^2=11\\a+b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(4-b\right)^2+2b^2=11\\a=4-b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}b=\dfrac{5}{3}\\b=1\end{matrix}\right.\\a=4-b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{5}{3}\\a=\dfrac{7}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}b=1\\a=3\end{matrix}\right.\end{matrix}\right.\)

TH1: \(a=\dfrac{7}{3};b=\dfrac{5}{3}\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{y}=\dfrac{7}{3}\\x-y=\dfrac{5}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+y=\dfrac{2}{3}\\x-y=\dfrac{5}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y^2-2y+3=0\left(vn\right)\\x-y=\dfrac{5}{3}\end{matrix}\right.\)

TH2:\(a=3;b=1\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{y}=3\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+y=2\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y^2-2y+1=0\\x-y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\) (thỏa mãn hệ)

Vậy hệ có nghiệm duy nhất (x;y)=(2;1).

\(1,\dfrac{x-1}{3}=x+1\\ \Leftrightarrow x-1=3x+3\\ \Leftrightarrow3x-x=3+1\\ \Leftrightarrow x=2\)

PT có tập nghiệm S = {2}

\(2,\sqrt{16x^2+8x+1}-2=x\\ \Leftrightarrow\sqrt{\left(4x+1\right)^2}-2=x\\\Leftrightarrow 4x+1-2=x\\ \Leftrightarrow4x-x=2-1\\ \Leftrightarrow x=\dfrac{1}{3}\)

PT có tập nghiệm S = {1/3}

\(3,\left\{{}\begin{matrix}2x+y=17\\x-2y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=17\\2x-4y=2\end{matrix}\right.\\ \Leftrightarrow\left(2x+y\right)-\left(2x-4y\right)=17-2\\ \Leftrightarrow5y=15\\ \Leftrightarrow y=3\\ \Leftrightarrow2x+3=17\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\)

PTHH có tập nghiệm (x; y) là (7; 3)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+y^4-4xy^3=1\\4x^2+2y^2-4xy=2\end{matrix}\right.\)

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\\6x+2y=-2\end{matrix}\right.-6x+12y=18}\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

\(\Rightarrow-4\left(x^3-y^3\right)=\left(5x^2-y^2\right)\left(16x-4y\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{7x}{4}\\y=-3x\end{matrix}\right.\)

Lần lượt thế vào \(y^2=5x^2+4\)...

b. Đề bài bất hợp lý, \(4x^2+y^4\) cần là \(4x^4+y^4\)

a) Ta có: \(\left\{{}\begin{matrix}49x+7y=-1\\-\dfrac{4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}98x+14y=-2\\-\dfrac{28}{3}x-14y=\dfrac{28}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{266}{3}x=\dfrac{22}{3}\\49x+7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\49\cdot\dfrac{11}{133}+7y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\7y=-1-\dfrac{77}{19}=-\dfrac{96}{19}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=13-4x=13-4\cdot\left(-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=21\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)