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\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{1}{4}-\dfrac{y}{3}+\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{x}{2}+\dfrac{5}{2}-\dfrac{y}{3}-\dfrac{7}{3}=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{5}{6}\\\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{25}{6}\end{matrix}\right.\) (vô lý)
Vậy HPT vô nghiệm
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.)
\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)
Thay vào (1) ta được :
\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}
\(b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\) (2)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = 1/x ; b = 1/y
Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{4};-3\) )}
c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)
ĐK xác định : x≠0 ; y ≠0
Áp dụng quy tác cộng đại số ta có :
\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)
ĐK xác định : x≠0 ; y≠0
áp dụng quy tắc cộng đại số ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)
Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}
e) ĐK xác định x≠0 ; y≠0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)
Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
a: ĐKXĐ: x<>-1 và y<>-1
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+2-2}{x+1}+\dfrac{y+1-1}{y+1}=2\\\dfrac{x+1-1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{2}{x+1}+1-\dfrac{1}{y+1}=2\\1-\dfrac{1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=2-3=-1\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=1+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=-1\\\dfrac{2}{x+1}-\dfrac{6}{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y-1}=3\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=-\dfrac{7}{3}\\\dfrac{1}{x+1}-3:\dfrac{-7}{3}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}+3\cdot\dfrac{3}{7}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}=2-\dfrac{9}{7}=\dfrac{5}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x+1=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: y<>0 và y<>-12
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y}-\dfrac{x}{y+12}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\cdot\dfrac{x}{y+12}=3\left(vôlý\right)\\\dfrac{x}{y}-\dfrac{x}{y+12}=1\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\varnothing\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)\)
mk lm 1 bài còn lại bn lm tương tự nha :
a) điều kiện xác định : \(x\ge0;y\ge1\)
đặc \(a=\sqrt{x};b=\sqrt{y-1}\)
\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}a+2b=5\\4a-b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
ta có : \(a=1\Rightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tmđk\right)\) ; \(b=2\Rightarrow\sqrt{y-1}=2\Leftrightarrow y=5\left(tmđk\right)\)
vậy phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;5\right)\)
b) bn đặc : \(a=\dfrac{1}{x};b=\dfrac{1}{y+12}\)
c) bn đặc : \(a=\dfrac{x}{x+1};b=\dfrac{y}{y+1}\)
nhớ điều kiện nha
<=>
2/(x+y)+3√(x-2)=7(*)
5/(x+y)-2√(x-2)=1-5/2=-3/2(**)
(*).5-(**).2
(15+4)√(x+2)=35+3=38
√(x-2)=2; x=6
2/(x+y)=1; => y=2-x=-4
(x,y)=(6,-4)
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: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\x+\dfrac{1}{x}-y-\dfrac{1}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=5\\y+\dfrac{1}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3.4}{x+y}+\dfrac{3.4}{x-y}=\dfrac{5}{2}\\\dfrac{4}{x+y}+\dfrac{2.4}{x+y}=\dfrac{4}{3}\end{matrix}\right.\\ Đặt.a=\dfrac{4}{x+y},b=\dfrac{4}{x-y}\\ \Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\a+2b=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\3a+6b=4\end{matrix}\right.\)
\(\Leftrightarrow\left(3a+6b\right)-3a-3b=4-\dfrac{5}{2}\\ \Leftrightarrow3b=\dfrac{3}{2}\Rightarrow b=\dfrac{1}{2}\Rightarrow a+2.\dfrac{1}{2}=\dfrac{4}{3}\\\Leftrightarrow a+1=\dfrac{4}{3}\Rightarrow a=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{4}{x+y}=\dfrac{1}{3}\\\dfrac{4}{x+y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=12\\x-y=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\)