Bài 2:

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

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

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

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1

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

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

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

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

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

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

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2-xy=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)

\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\)

\(\Leftrightarrow x^4-5x^2=4=0\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Đặt \(\left(x-2\right)^2=a;\dfrac{1}{\sqrt{y+5}}=b\)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}2a+b=3\\a-2b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

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

⚠

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.)

a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)

ĐK: \(x>\dfrac{1}{4};y>-\dfrac{1}{4}\), đặt \(a=\dfrac{1}{\sqrt{12x-3}};b=\dfrac{1}{\sqrt{4y+1}}\)với a,b>0

khi đó, ta có hệ phương mới \(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}10a+5b=1\\7a+8b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}80a+40b=8\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45a=3\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35a+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\35.\dfrac{1}{15}+40b=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{15}\\b=\dfrac{1}{15}\end{matrix}\right.\)

thay \(\dfrac{1}{\sqrt{12x-3}}=a\) hay \(\dfrac{1}{\sqrt{12x-3}}=\dfrac{1}{15}\Rightarrow\sqrt{12x-3}=15\Leftrightarrow12x-3=225\Leftrightarrow12x=228\Leftrightarrow x=19\left(TMĐK\right)\) thay \(\dfrac{1}{\sqrt{4y+1}}=b\) hay

\(\dfrac{1}{\sqrt{4y+1}}=\dfrac{1}{15}\Rightarrow\sqrt{4y+1}=15\Leftrightarrow4y+1=225\Leftrightarrow4y=224\Leftrightarrow y=56\left(TMĐK\right)\)

Vậy (x;y)=(9;56) là nghiệm duy nhất của hệ phương trình đã cho.

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

ĐK: x,y#0, khi đó \(\dfrac{1}{x}+\dfrac{1}{y}=4\Rightarrow x+y=4xy\)

Do đó \(x\left(1+4y\right)+y=2\Leftrightarrow x+4xy+y=2\Leftrightarrow x+x+y+y=2\Leftrightarrow2\left(x+y\right)=2\Leftrightarrow x+y=1\)

Mà \(4xy=x+y\Leftrightarrow4xy=1\Leftrightarrow xy=\dfrac{1}{4}\)

Vậy \(x+y=1;xy=\dfrac{1}{4}\)

Do đó x,y là nghiệm của phương trình:

\(t^2-t+\dfrac{1}{4}=0\)

\(\Delta=b^2-4ac=1-4.1.\dfrac{1}{4}=0\)

Phương trình có nghiêm kép \(x_1=x_2=-\dfrac{b}{2a}=-\dfrac{-1}{2}=\dfrac{1}{2}\)

\(\Rightarrow x=y=\dfrac{1}{2}\left(nhận\right)\)

Vậy (x;y)=\(\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) là nghiệm duy nhất của hệ phương trình đã cho.

c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)

Trừ vế đối vế hai phương trình, ta được:

\(x^2+x+1-y^2-y-1=3y-3x\\ \Leftrightarrow x^2-y^2+4x+4y=0\\ \Leftrightarrow\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x+y+4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x+y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-x-4\end{matrix}\right.\)

+Với x=y thế vào \(x^2+x+1=3y\) ta được

\(x^2+x+1=3x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Do đó (x;y)=(1;1) là một nghiệm của hệ phương trình đã cho.

+Với y=-x-4 thế vào \(x^2+x+1=3y\) ta được

\(x^2+x+1=3\left(-x-4\right)\Leftrightarrow x^2+4x+13=0\Leftrightarrow\left(x+2\right)^2+9=0\)(*)

Mặt khác \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+9\ge0\Rightarrow\left(x+2\right)^2\ge-9>0\), do đó phương trình (*) vô nghiệm

Vậy (x;y)=(1;1) là nghiệm duy nhất của hệ phương trình đã cho.

ĐKXĐ : x;y \(\ne0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{1}{x}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=\dfrac{1}{9}\end{matrix}\right.\)

ĐK: `x ne 2; y ne -1`

Đặt `{a=(1/(x-2)),(b=1/(y+1)):}`

Có: `{(2a+b=3),(4a-3b=1):}`

`<=>{(4a+2b=6),(4a-3b=1):}`

`<=>{(2a+b=3),(5b=5):}`

`<=>{(2a+1=3),(b=1):}`

`<=>{(a=1),(b=1):}`

``

`=>{(1/(x-2)=1),(1/(y+1)=1):}`

`<=>{(x-2=1),(y+1=1):}`

`<=>{(x=3),(y=0):}` (TM)

``

Vậy `(x;y)=(3;0)`.