2x + 6y + 2z = 60 ( 1 )
x : y : z = 1 : 2 : 3 ( 2 )
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f ) x + y = x . y = x : y
Ta có :
\(x+y=xy\Rightarrow x=xy-y=y\cdot\left(x-1\right)\\ \Rightarrow x:y=x-1\)
Mặt khác , x : y = x + y ( gt )
\(\Rightarrow x-1=x+y\\ \Rightarrow x-x=1+y\\ \Rightarrow1+y=0\\ \Rightarrow y=-1\)
\(+)x=\left(x-1\right)\cdot y\\ \Rightarrow x=\left(x-1\right)\cdot\left(-1\right)\\ \Rightarrow x=-x+1\\ \Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
Vậy x = \(\dfrac{1}{2},y=-1\)
1) 2x + 2y - x(x+y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2/ 5x2 - 5xy -10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10(x - y)
3/ 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
1) 2x + 2y - x(x + y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2) 5x2 - 5xy - 10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10)(x - y)
= 5(x - 2)(x - y)
3) 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
4) 2x2 + 2y2 - x2z + z - y2z - 2
= 2(x2 + y2 - z(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2)
5) x2 + xy - 5x - 5y
= x(x + y) - 5(x + y)
= (x - 5)(x + y)
6) x(2x - 7) - 4x + 14
= x(2x - 7) - 2(2x - 7)
= (x - 2)(2x - 7)
7)x2 - 3x + xy - 3y
= x(x + y) - 3(x + y)
= (x - 3)(x + y)
\(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-2z\left(x-y\right)+z+\left(x^2-8x+16\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\left(x-y-z\right)^2+\left(x-4\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y-z=0\\x-4=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=1\\x=4\\y=3\end{cases}}\)
Vậy \(x=4\), \(y=3\), \(z=1\)
a) Hai mặt phẳng cắt nhau, vì 1: 2: (-1) ≠ 2: 3: (-7)
b) Hai mặt phẳng cắt nhau, vì: 1: (-2): 1 ≠ 2: (-1): 4
c) Hai mặt phẳng song song, vì: 1/2=1/2=1/2 ≠ -1/3
d) Hai mạt phẳng cắt nhau, vì: 3: (-2): 3 ≠ 9: (-6): (-9)
e) Hai mặt phẳng trung nhau, vì: 1/10=-1/(-10)=2/20=-4/(-40).
#rin
\(\frac{2}{x+y+z}=\frac{x}{2y+2z+1}=\frac{y}{2x+2z+1}=\frac{z}{2x+2y-2}=\frac{x+y+z}{4\left(x+y+z\right)}=\frac{1}{4}\)
\(\Rightarrow\hept{\begin{cases}2y+2z+1=4x\\2x+2z+1=4y\\x+y+z=8\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=\frac{17}{6}\\z=\frac{7}{3}\end{cases}}\)
Ta có:
\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)
Không tồn tại x,y,z thỏa mãn đề bài