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

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

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

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

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)

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

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)

hỏi trước tí, bạn biết giải cái hệ này chứ?

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

ba cái đồ êu!!

câu số 6 (con số của quỷ sa tăng :v)

đặt \(\left\{{}\begin{matrix}a=\left|x\right|\\b=\left|y\right|\end{matrix}\right.\) (a,b >/ 0)

hpt trở thành : \(\left\{{}\begin{matrix}a+4b=18\\3a+b=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x\right|=2\\\left|y\right|=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}y=4\\y=-4\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có các ng (x;y) là: (có 4 nghiệm tự kết luận)

1, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\) (I) (ĐKXĐ: x, y \(\ne\)0)

Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\)

Hệ pt (I) trở thành :

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\) \(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-7b=\dfrac{-1}{3}\\a+b=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a+\dfrac{1}{21}=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\left(tm\right)\\a=\dfrac{1}{28}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{28}\\\dfrac{1}{y}=\dfrac{1}{21}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

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

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

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

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

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

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

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

-> \(\left\{{}\begin{matrix}y=1\\\sqrt{x+1}=1\end{matrix}\right.\)

-> \(\left\{{}\begin{matrix}y=1\\x=0\end{matrix}\right.\)

Vậy .........

1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)

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

2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)

\(=3\sqrt{x}\)

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

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

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

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

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

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)

=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75

=>x=7; y=5

b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)

=>4x+9y=8 và -8x+3y=5

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

c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)

=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5

=>2x-3y=-5,5 và 3x-2y=-4,5

=>x=-1/2; y=3/2

e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)

=>\(x=\sqrt{2};y=\sqrt{3}\)

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

Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x-1}\\b=\dfrac{y}{y+1}\end{matrix}\right.\left(a\ge0\right)\)

hệ pt trở thành \(\left\{{}\begin{matrix}3a-b=1\\a+2b=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6a-2b=2\left(1\right)\\a+2b=5\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)+\left(2\right)\Rightarrow7a=7\Rightarrow a=1\Rightarrow b=2\)

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

\(\sqrt{2x-1}=1\Rightarrow x=1\)

\(\dfrac{y}{y+1}=2\Rightarrow2y+2=y\Rightarrow y=-2\)

Vậy hệ có bộ nghiệm (x,y) là \(\left(1,-2\right)\)