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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 \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
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: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}49x+7y=-1\\-\dfrac{4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}98x+14y=-2\\-\dfrac{28}{3}x-14y=\dfrac{28}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{266}{3}x=\dfrac{22}{3}\\49x+7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\49\cdot\dfrac{11}{133}+7y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\7y=-1-\dfrac{77}{19}=-\dfrac{96}{19}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=13-4x=13-4\cdot\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=21\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại
Với pt sau:
Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm
Với \(x;y;z\ne0\)
Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3
Do đó:
\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)
Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)
a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)
=>x=2 và y=-2
Giải hệ sau :
Câu a :
\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy ...........................
Câu b :
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
9: \(\left\{{}\begin{matrix}3x-2=y\\2x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=2\\2x+3y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)
\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)
\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)
Còn lại tương tự
a: \(\left\{{}\begin{matrix}3x-2y=1\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-4y=2\\2x+4y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x=5\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\2y=3x-1=\dfrac{15}{8}-1=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=\dfrac{7}{16}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}4x-3y=1\\-x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-3y=1\\-4x+8y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1+2y=-1+2=1\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{4}{3}y=1\\\dfrac{1}{2}x-\dfrac{3}{4}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=3\\2x-3y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{41}{14}\\y=-\dfrac{5}{7}\end{matrix}\right.\)