giải hệ sau \(\left\{{}\begin{matrix}x=3y^3+2y^2+y\\y=3z^3+2z^2+z\\z=3x^3+2x^2+x\end{matrix}\right.\)
Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}x+2y-3z=2\\2x+7y+z=5\\-3x+3y-2z=-7\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}-x-3y+4z=3\\3x+4y-2z=5\\2x+y+2z=4\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x+2y-3z=2\\2x+7y+z=5\\-3x+3y-2z=-7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2y-3z=2\\3y+7z=1\\-32z=-4\end{matrix}\right.\)
Đáp số : \(\left(x,y,z\right)=\left(\dfrac{55}{24},\dfrac{1}{24},\dfrac{1}{8}\right)\)
b) \(\left\{{}\begin{matrix}-x-3y+4z=3\\3x+4y-2z=5\\2x+y+2z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-3y+4z=3\\-5y+10z=14\\-5y+10z=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-3y+4z=3\\-5y+10z=14\\0y+0z=-4\end{matrix}\right.\)
Phương trình cuối vô nghiệm, suy ra hệ phương trình đã cho vô nghiệm
Giải các hệ pt, bất pt sau:
a, \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}2x-3y\ge2\\3x+2y< 4\\x-2y\ge5\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}2x-2y+z=3\\2x+y-2z=-3\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-4y+2z=6\\8x+4y-8z=-3\\3x-4y-z=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12x-6z=3\\11x-9z=1\\3x-4y-z=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\z=\dfrac{1}{2}\\4y=3x-z-4=\dfrac{3}{2}-\dfrac{1}{2}-4=1-4=-3\end{matrix}\right.\)
=>x=1/2;z=1/2;y=-3/4
Tìm x,y,z thỏa mãn hệ sau:
\(\left\{{}\begin{matrix}x^3-3x-2=2-y\\y^3-3y-2=4-2z\\z^3-3z-2=6-3x\end{matrix}\right.\)
Giải hệ phương trình :
a) \(\left\{{}\begin{matrix}x^2+y^2=1\\x^2+y^2=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=2014\\\dfrac{1}{3x+2y}+\dfrac{1}{3y+2z}+\dfrac{1}{3z+2x}=\dfrac{1}{x+2y+3z}+\dfrac{1}{y+2x+3x}+\dfrac{1}{z+2x+3y}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3x^2+2y+4=2z\left(x+3\right)\\3y^2+2z+4=2x\left(y+3\right)\\3z^2+2x+4=2y\left(z+3\right)\end{matrix}\right.\)
Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}x-2y+z=\\2x-y+3z=18\\-3x+3y+2z=-9\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+z=7\\3x-2y+2z=5\\4x-y+3z=10\end{matrix}\right.\)
b) Đặt \(\left\{{}\begin{matrix}x+y+z=7\left(1\right)\\3x-2y+2z=5\left(2\right)\\4x-y+3z=10\left(3\right)\end{matrix}\right.\)
Cộng \(\left(1\right)+\left(2\right)\) ta có: \(4x-y+3z=12\). (4)
Từ (3) và (4): \(\left\{{}\begin{matrix}4x-y+3z=12\\4x-y+3z=10\end{matrix}\right.\) (vô nghiệm).
Vậy hệ phương trình vô nghiệm.
Giải các hệ phương trình :
a. \(\left\{{}\begin{matrix}x+3y+2z=8\\2x+2y+z=6\\3x+y+z=6\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x-3y+2z=-7\\-2x+4y+3z=8\\3x+y-z=5\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x+3y+2z=8\\2x+2y+z=6\\3x+y+z=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x-3y+2z=-7\\-2x+4y+3z=8\\3x+y-z=5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{14}\\y=\dfrac{5}{2}\\z=-\dfrac{1}{7}\end{matrix}\right.\)
a) Đặt \(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\3x+y+z=6\left(3\right)\end{matrix}\right.\)
Cộng \(\left(2\right)+\left(3\right)\) ta có:\(\left\{{}\begin{matrix}x+3y+2z=8\left(1\right)\\2x+2y+z=6\left(2\right)\\5x+3y+2z=12\left(4\right)\end{matrix}\right.\)
Trừ \(\left(4\right)-\left(1\right)\) ta được: \(4x=4\Leftrightarrow x=1\).
Thay vào hệ phương trình ta được:
\(\left\{{}\begin{matrix}1+3y+2z=8\\2.1+2y+z=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=2\end{matrix}\right.\).
Vậy hệ phương trình có nghiệm: \(\left\{{}\begin{matrix}x=1\\y=1\\z=2\end{matrix}\right.\).
b) Đặt \(\left\{{}\begin{matrix}x-3y+2z=-7\left(1\right)\\-2x+4y+3z=8\left(2\right)\\3x+y-z=5\left(3\right)\end{matrix}\right.\)
Cộng \(\left(1\right)-\left(2\right)\) ta được: \(3x-7y-z=-15\left(4\right)\)
Lấy \(\left(3\right)-\left(4\right)\) ta được: \(8y=20\Leftrightarrow y=\dfrac{5}{2}\).
Thay \(y=\dfrac{5}{2}\) vào hệ phương trình ta có:
\(\left\{{}\begin{matrix}x-3.\dfrac{5}{2}+2z=-7\\-2x+4.\dfrac{5}{2}+3z=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{14}\\z=-\dfrac{1}{7}\end{matrix}\right.\).
Vậy hệ có nghiệm là: \(\left\{{}\begin{matrix}x=\dfrac{11}{14}\\y=\dfrac{5}{2}\\z=\dfrac{-1}{7}\end{matrix}\right.\)
giải hệ phương trình: \(\left\{{}\begin{matrix}x^3-3x=4-y\\y^3-3y=6-2z\\z^3-3z=8-3x\end{matrix}\right.\)
Giải các hệ phương trình sau bằng máy tính bỏ túi (làm tròn kết quả dến chữ số thập phân thứ hai)
a. \(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}-2x+3y=5\\5x+2y=4\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}2x-3y+4z=-5\\-4x+5y-z=6\\3x+4y-3z=7\end{matrix}\right.\)
d. \(\left\{{}\begin{matrix}-x+2y-3z=2\\2x+y+2z=-3\\-2x-3y+z=5\end{matrix}\right.\)
a. \(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)
\(x=\dfrac{2}{41}\) ; \(y=\dfrac{-48}{41}\)
b. \(\left\{{}\begin{matrix}\text{−2x+3y=5}\\5x+2y=4\end{matrix}\right.\)
\(x=\dfrac{2}{19};y=\dfrac{33}{19}\)
c.\(\left\{{}\begin{matrix}\text{2x−3y+4z=−5}\\-4x+5y-z=6\\3x+4y-3z=7\end{matrix}\right.\)
\(x=\dfrac{22}{101};y=\dfrac{131}{101};z=\dfrac{-39}{101}\)
d. \(\left\{{}\begin{matrix}\text{− x + 2 y − 3 z = 2}\\2x+y+2z=-3\\-2x-3y+z=5\end{matrix}\right.\)
\(x=-4;y=\dfrac{11}{7};z=\dfrac{12}{7}\)
a)x=0,05 ; y=-1,17
b.x=0,11 ; y=1,74
c.x=0,22 ;y=1,29 z=-0.39
d.x=-4 y=1,57 z=1,71
a,\(\left\{{}\begin{matrix}3x-5y=6\\4x+7y=-8\end{matrix}\right.\)
x=\(\dfrac{2}{41}=0,05\) ; y=\(\dfrac{-48}{41}=-1,17\)
b,\(\left\{{}\begin{matrix}-2x+3y=5\\5x+2y=4\end{matrix}\right.\)
x=\(\dfrac{2}{19}=0,11\) ; y=\(\dfrac{33}{19}=1,74\)
c,\(\left\{{}\begin{matrix}2x-3y+4z=-5\\-4x+5y-z=6\\3x+4y-3z=2\end{matrix}\right.\)
x=\(\dfrac{22}{101}=0,22\) ;y=\(\dfrac{131}{101}=1,29\) ; z=\(\dfrac{-39}{101}=-0,39\)
d,\(\left\{{}\begin{matrix}-x+2y-3z=2\\2x+y+2z=-3\\-2x-3y+z=5\end{matrix}\right.\)
x=\(-4\) ; y=\(\dfrac{11}{7}=1,57\) ; z=\(\dfrac{12}{7}=1,71\)