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

a)\(\Leftrightarrow\left\{{}\begin{matrix}25x+15y=40xy\left(1\right)\\24x+16y=40xy\left(2\right)\end{matrix}\right.\)

Lấy (1) trừ (2), ta được: x-y=0\(\Leftrightarrow x=y\)

Thay vào 5x+3y=8xy ta được: \(5x+3x=8x^2\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\).\(\Rightarrow\left[{}\begin{matrix}x=y=0\\x=y=1\end{matrix}\right.\)

Vậy hpt có nghiệm (0;0);(1;1).

b)\(\Leftrightarrow\left\{{}\begin{matrix}-5x+5y=5xy\left(1\right)\\4x+3y=5xy\left(2\right)\end{matrix}\right.\)

Lấy (2) trừ (1) ta được: 9x-2y=0 \(\Leftrightarrow y=\dfrac{9x}{2}\)

Thay vào -x+y=xy ta được: \(-x+\dfrac{9x}{2}=x^2\)

\(\Leftrightarrow-2x+9x=2x^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{7}{2}\left(KTM\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=0\left(TM\right)\\y=\dfrac{63}{4}\left(KTM\right)\end{matrix}\right.\)

Vậy hpt có nghiệm (0;0).

c) Từ 2x-y=5\(\Rightarrow y=2x-5\)

Thay vào \(\left(x+y+2\right)\left(x+2y-5\right)=0\), ta được:

\(\left(3x-3\right)\left(5x-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=5\left(KTM\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=1\left(TM\right)\\y=5\left(KTM\right)\end{matrix}\right.\)

Vậy hpt có nghiệm (3;1).

a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)

cộng vế pt (1) và (2), ta được:

\(3x^2+4xy+y^2=4+4x\Leftrightarrow\left(2x+y\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(3x+y-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=2-3x\end{matrix}\right.\)

thay từng trường hợp vào pt (1) giải tiếp

a. ĐKXĐ: ..

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)

b.

Thế pt trên xuống dưới:

\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)

\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)

Thế vào pt đầu ...

Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều

\(2x^2+30xy=5\left(x+5y\right)\sqrt{5xy}-50y^2\)\(\left(đk:x;y\ge0\right)\)

\(\Leftrightarrow2x^2+30xy-5\left(x+5y\right)\sqrt{5xy}+50y^2=0\left(1\right)\)

\(đặt:\sqrt{5xy}=b\ge0\Rightarrow5xy=b^2\Rightarrow10xy=2b^2\)

\(x+5y=a\ge0\Rightarrow x^2+10xy+25y^2=â^2\)

\(\Rightarrow2a^2=2x^2+20xy+50y^2\)

\(\Leftrightarrow\left(1\right)\Leftrightarrow2a^2+2b^2-5ab=0\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}b=2a\left(2\right)\\a=2b\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow\sqrt{5xy}=2x+10y\Leftrightarrow4x^2+35xy+100y^2=0\left(4\right)\)

\(với:y=0\) \(ko\) \(là\) \(nghiệm\)

\(với:y\ne0\Rightarrow\left(4\right)\Leftrightarrow4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100=0\)\(\left(vô-lí\right)\)

\(do:4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100>0\)

\(\left(3\right)\Rightarrow x+5y=2\sqrt{5xy}\Leftrightarrow x^2+10xy+25y^2=20xy\Leftrightarrow x^2-10xy+25y^2=0\Leftrightarrow\left(x-5y\right)^2=0\Leftrightarrow x=5y\)

\(thay:x=5y\) \(vào:2x^2+y^2=51\Rightarrow2\left(5y\right)^2+y^2-51=0\Leftrightarrow51y^2-51=0\Leftrightarrow\left[{}\begin{matrix}y=1\left(tm\right)\Rightarrow x=5\left(tm\right)\\y=-1\left(loại\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}6x+6y=5xy(1)\\\dfrac{4}{x}-\dfrac{3}{y}=1\end{matrix}\right.\)
Chia 2 vế cho xy thì (1)(vì `x,y ne 0`)
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{4}{x}-\dfrac{3}{y}=1\\\end{cases}$
`<=>` $\begin{cases}\dfrac6x+\dfrac6y=5\\\dfrac{8}{x}-\dfrac{6}{y}=2\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}\dfrac{14}{x}=7\\\dfrac6x+\dfrac6y=5\\\end{cases}$
`<=>` $\begin{cases}x=2\\y=3\\\end{cases}$
Vậy HPT có nghiệm (x,y)=(2,3)

\(\left\{{}\begin{matrix}3x-3y=5\\5x+2y=23\end{matrix}\right.< =>\left\{{}\begin{matrix}6x-6y=10\\15x+6y=69\end{matrix}\right.< =>\left\{{}\begin{matrix}21x=79\\3x-3y=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=\dfrac{79}{21}\\y=\dfrac{44}{21}\end{matrix}\right.\)

vậy hệ pt có nghiệm (x,y)=(\(\dfrac{79}{21};\dfrac{44}{21}\))