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DO
2
20 tháng 12 2019
Ta có: x + y = 9
x + z = 15
y + z = 12
=> x + y + x + z + y + z = 9 + 15 + 12
<=> 2( x + y + z ) = 36
<=> x + y + z = 18
=> x = 18 - 12 = 6
y = 18 - 15 = 3
z = 18 - 9 = 9
CC
20 tháng 12 2019
Từ \(x+y=9\), \(x+z=15\), \(y+z=12\)
\(\Rightarrow x+y+x+z+y+z=9+15+12\)\(\Rightarrow2\left(x+y+z\right)=36\)
\(\Rightarrow x+y+z=18\)
Ta có: \(x+y=9\)\(\Rightarrow z=18-9=9\)
\(x+z=15\)\(\Rightarrow y=18-15=3\)
\(y+z=12\)\(\Rightarrow x=18-12=6\)
Vậy \(x=6\), \(y=3\), \(z=9\)
LV
2
18 tháng 4 2020
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BT
30 tháng 11 2017
Ta có: x+y-z=y+z-x <=> 2x=2z => x=z
Lại có: y+z-x=z+x-y <=> 2x=2y => x=y
=> x=y=z
Do x+y-z=xyz => x=x3 => x(x2-1)=0 <=> x(x-1)(x+1)=0
=> x1=y1=z1=0 ; x2=y2=z2=1 ; x3=y3=z3=-1
6 tháng 10 2018
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)
Thay số vào tính được \(xy+yz+xz=12\)
Ta có: \(x^2+y^2+z^2=xy+yz+xz\left(=12\right)\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Từ đó được \(x=y=z\)
Mà \(x+y+z=6\Rightarrow x=y=z=2\)
Chúc bạn học tốt.
VD
26 tháng 6 2016
bài này hoàn toàn có thể cosi dù đề bài chưa cho dương hoac su dung bunhia ngc ( thi ko can quan tam duong hay am)
19 tháng 12 2016
Đặt: \(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}=M\)
Ta có:
\(M\cdot\frac{z}{x-y}=1+\frac{z}{x-y}\cdot\left(\frac{y-z}{x}+\frac{z-x}{y}\right)=1+\frac{z}{x-y}\cdot\frac{y^2-yz+xz-x^2}{xy}\)
\(=1+\frac{z}{x-y}\cdot\frac{\left(x-y\right)\left(z-x-y\right)}{xy}=1+\frac{2z^2}{xyz}=1+\frac{2z^3}{xyz}\) (1)
Tương tự ta cũng có:
\(M\cdot\frac{x}{y-z}=1+\frac{2x^3}{xyz}\) (2)
\(M\cdot\frac{y}{z-x}=1+\frac{2y^3}{xyz}\) (3)
Từ (1);(2);(3) suy ra
\(M\cdot\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)=3+\frac{2\left(x^3+y^3+z^3\right)}{xyz}\)
Mà \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
Nên:
\(M\cdot\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)=3+\frac{2\cdot3xyz}{xyz}=9\)
=>đpcm
Giải:
Ta có:
\(\left(x+y\right)\left(x+z\right)=15\); \(\left(y+z\right)\left(y+x\right)=18\); \(\left(z+x\right)\left(z+y\right)=30\)
\(\Leftrightarrow\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2=15.18.30\)
\(\Leftrightarrow\left(\left(x+y\right)\left(y+z\right)\left(z+x\right)\right)^2=8100\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=90\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{90}{30}=3\\y+z=\dfrac{90}{15}=6\\z+x=\dfrac{90}{18}=5\end{matrix}\right.\)
\(\Leftrightarrow2\left(x+y+z\right)=3+6+5=14\)
\(\Leftrightarrow x+y+z=7\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7-6=1\\y=7-5=2\\z=7-3=4\end{matrix}\right.\)
Vậy ...
Ta có:
\(\left\{{}\begin{matrix}\left(x+y\right)\left(z+x\right)=15\\\left(x+y\right)\left(y+z\right)=18\\\left(y+z\right)\left(z+x\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2=8100\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=90\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\y+z=6\\z+x=5\end{matrix}\right.\)
\(\Leftrightarrow2\left(x+y+z\right)=14\)
\(\Leftrightarrow x+y+z=7\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\\z=4\end{matrix}\right.\)