a, Cho \(A=\sqrt{2012^2+2012^2.2013^2+2013^2}\). CMR A là 1 STN
b, Giải hệ \(\left\{{}\begin{matrix}x^2+\dfrac{1}{y^2}+\dfrac{x}{y}=3\\x+\dfrac{1}{y}+\dfrac{x}{y}=3\end{matrix}\right.\)
Bài 1: Giải các hệ PT
a) \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\) c) \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+1\right)=\dfrac{1}{2}xy+5\\\dfrac{1}{3}\left(x-3\right)\left(y-5\right)=\dfrac{1}{3}xy-\dfrac{4}{3}\end{matrix}\right.\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}+\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=4\end{matrix}\right.\)
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: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
Bìa 1: Gải các hệ phương trình:
a) \(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\) b)\(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
Bài 2: Gải các hệ phương trình:
a) \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)
Bài 3: Gải các hệ phương trình:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}\dfrac{1}{2x+y}+\dfrac{1}{x-2y}=\dfrac{5}{8}\\\dfrac{1}{2x+y}-\dfrac{1}{x-2y}=\dfrac{3}{8}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\left|x-1\right|+\left|y+2\right|=2\\4\left|x-1\right|+3\left|y+2\right|=7\end{matrix}\right.\)
Bài 4: Cho hệ phương trình \(\left\{{}\begin{matrix}\left(3a-2\right)x+2\left(2b+1\right)y=30\\\left(a+2\right)x-2\left(3b-1\right)y=-20\end{matrix}\right.\) Tìm các giá trị của a,b để hệ phương trình có nghiệm (3;-1)
cảm ơn mn trước ạ ! hehe
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
Giải các hệ phương trình sau:
1, \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\dfrac{1}{x-5}+\dfrac{6}{\sqrt{y}-2}=2\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x+7}}-\dfrac{4}{\sqrt{y-6}}=\dfrac{5}{3}\\\dfrac{5}{\sqrt{x+7}}+\dfrac{3}{\sqrt{y-6}}=\dfrac{13}{6}\end{matrix}\right.\)
Tìm điều kiện giúp mình nhé!
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
Giải các hệ phương trình sau bằng cách đặt ẩn số phụ:
1) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}x^2+y^2=13\\3x^2-2y^2=-6\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=16\\2\sqrt{x}-3\sqrt{y}=-11\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}|x|+4|y|=18\\3|x|+|y|=10\end{matrix}\right.\)
GIẢI GIÚP MÌNH VỚI M.N
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 :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{12}\\8a+15b=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}8a+8b=\dfrac{2}{3}\\8a+15b=1\end{matrix}\right.\) \(\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-7b=\dfrac{-1}{3}\\a+b=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a+\dfrac{1}{21}=\dfrac{1}{12}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}b=\dfrac{1}{21}\left(tm\right)\\a=\dfrac{1}{28}\left(tm\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{28}\\\dfrac{1}{y}=\dfrac{1}{21}\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
1. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}x-\sqrt{3y}=\sqrt{3}\\\sqrt{3x}+y=7\end{matrix}\right.\)
2. Giải các hpt sau:
a, \(\left\{{}\begin{matrix}2-\left(x-y\right)-3\left(x+y\right)=5\\3\left(x-y\right)+5\left(x+y\right)=-2\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}\dfrac{2}{x-2}+\dfrac{2}{y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{y-1}=1\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{9}+\dfrac{y}{27}=2\dfrac{8}{9}\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2=15}\end{matrix}\right.\)
3. Cho hpt \(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\)
a, Giải hpt khi m=\(\sqrt{2}\)
b, tìm giá trị của m để hpt có nghiệm duy nhất thỏa mãn: x+y>0
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
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}y=2\sqrt{x-1}\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}xy-\dfrac{x}{y}=9.6\\xy-\dfrac{y}{x}=7.5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)