6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)

7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)

8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y

(Các câu khác tương tự nhé.)

Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)

Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)

=> hpy vô nghiệm

c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)

Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt

\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)

với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)

đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !

b) Ta có hpt <=> \(\left\{{}\begin{matrix}2\sqrt{x}-3y+2=-4z^2\\2\sqrt{3x}+4y-2=6z^2\\-3\sqrt{x}+y-4=-2z^2\end{matrix}\right.\)

cộng 3 vế của 3 pt, ta có \(\left(2\sqrt{3}-1\right)\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{2\sqrt{3}-1}\Leftrightarrow x=\dfrac{16}{\left(2\sqrt{3}-1\right)^2}\)

đến đây thay căn(x)=...vào và đặt z^2=m, ta sẽ ra 1 hệ mới chỉ có 2 ẩn y và m bậc 1 , lát thế vào sẽ ra bậc 2 thì dễ rồi !

a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)

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

c: ĐKXĐ: y>2

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

=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)

\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)

\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}\dfrac{-5x+2y}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(-5x+2y\right)+60=3\left(y+27\right)-24x\\7\left(x+1\right)+21y=3\left(6y-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y+60=3y+81-24x\\7x+7+21y=18y-15x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y-3y+24x=21\\7x+21y-18y+15x=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+15y=63\\110x+15y=-35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-98x=98\\4x+5y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\5y=21-4x=21+4=25\end{matrix}\right.\)

=>x=-1 và y=5

b: \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+3x+2y+6-xy=100\\xy-\left(xy-2x-2y+4\right)=64\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y=94\\2x+2y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=34\\2x+2y=60\end{matrix}\right.\)

=>x=34 và y=-4

c: \(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-x+20y-20=xy\\xy+x-10y-10=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+20y=20\\x-10y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=30\\x-10y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=10y+10=30+10=40\end{matrix}\right.\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-2y\\x< >-\dfrac{y}{2}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}\dfrac{4}{x+2y}+\dfrac{2}{2x+y}=6\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2x+y}=5\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+y=1\\\dfrac{4}{x+2y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=1\\x+2y=1\end{matrix}\right.\)

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

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

e: ĐKXĐ: x<>-1 và y<>-4

\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\left(nhận\right)\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?