Cho \(x+y+z=0\) Rút gọn : \(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Cho x+y+z=0 Rút gọn:\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Ta có: \(x+y+z=0\Rightarrow\hept{\begin{cases}-x=y-z\\-y=z-x\\-z=x-y\end{cases}}\)
Mà \(x^2=\left(-x\right)^2;y^2=\left(-y\right)^2;z^2=\left(-z\right)^2\)
Thế vào biểu thức, ta được:
\(\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1\)
bn ơi bài lm của
BÀI LÀM.
\(x+y+z=0\)
\(\Leftrightarrow\)\(\left(x+y+z\right)^2=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Leftrightarrow\)\(x^2+y^2+z^2=-2zy-2yz-2zx\)
Ta có: \(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)
\(=y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2\)
\(=2x^2+2y^2+2z^2-2xy-2yz-2xz\)
\(=2x^2+2y^2+2z^2+x^2+y^2+z^2\) (thay -2y - 2yz - 2zx = x^2 +y^2 +z^2)
\(=3\left(x^2+y^2+z^2\right)\)
Vậy \(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
CẦN GẤP!!
Cho x+y+z=0 xà x,y,z khác 0 rút gọn
a)P=\(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b)Q=\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)
\(\Rightarrow P=\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left[-\left(y+z\right)\right]^2+\left[-\left(z+x\right)\right]^2+\left[-\left(x+y\right)\right]^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left(y+z\right)^2+\left(z+x\right)^2\left(x+y\right)^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{-\left[\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=-1\)
Mik mới biết làm câu a thôi còn câu b thì từ từ mik nghĩ đã nhé @-@
Chúc bn học giỏi nhoa!!!
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
ta có:x+y+z=0⇒x+y=-z⇔(x+y)2=z2⇔x2+2xy+y2-z2=0
⇒x2+y2-z2=-2xy(1)
CMTT:⇒y2+z2-x2=-2yz(2) và z2+x2-y2=-2xz(3)
Thay (1)(2)(3) vào B,ta có.B=-(2xy.2yz.2xz)/16xyz=-xyz/2
1rút gọn\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)biết rằng x+y+z=0
2 rút gọn các phân thức
a,\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
b,\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Rút gọn \(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\) biết x+y+z=0
Ta có: \(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Rightarrow x^2+y^2+z^2=-2.\left(xy+yz+zx\right)\)
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{y^2+z^2+z^2+x^2+x^2+y^2-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{2.\left[-2.\left(xy+yz+zx\right)\right]-2.\left(xy+yz+zx\right)}\)
\(=\frac{-2.\left(xy+yz+zx\right)}{-6.\left(xy+yz+zx\right)}\)
\(=\frac{1}{3}\left(xy+yz+zx\ne0\right)\)
Tham khảo nhé~
cho x + y + z = 0.Rút gọn
a)\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
b)\(\frac{2x^2y+2xy^2}{x^2+y^2-z^2}\)
c)\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xy^2}\)
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
\(x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-2\left(xy+yz+zx\right)\right]-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}\)
\(=\frac{1}{3}\)
Ta có: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\)
\(x^2+2xy+y^2=z^2\)
\(x^2+y^2-z^2=-2xy\)
\(\frac{2x^2y+2xy^2}{x^2+y^2-z^2}\)
\(=\frac{2xy\left(x+y\right)}{-2xy}\)
\(=\frac{-2xyz}{-2xy}\)
\(=z\)
Ta có: \(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2=z^2\\\left(x+z\right)^2=y^2\\\left(y+z\right)^2=x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+2xy+y^2=z^2\\x^2+2xz+z^2=y^2\\y^2+2yz+z^2=x^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2-z^2=-2xy\\x^2+z^2-y^2=-2xz\\y^2+z^2-x^2=-2yz\end{cases}}\)
\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xy^2}\)
\(=\frac{\left(-2xy\right).\left(-2yz\right).\left(-2xz\right)}{16xy^2}\)
\(=\frac{-8x^2y^2z^2}{16xy^2}\)
\(=\frac{-xz^2}{2}\left(x,y\ne0\right)\)
Với x + y + z = 0. Rút gọn:
\(A=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Ta có \(x+y+z=0\Rightarrow x+y=-z\Rightarrow x-y=z\Rightarrow\left(x-y\right)^2=z^2\)
\(x+y+z=0\Rightarrow x+z=-y\Rightarrow z-x=y\Rightarrow\left(z-x\right)^2=y^2\)
\(x+y+z=0\Rightarrow y+z=-x\Rightarrow y-z=x\Rightarrow\left(y-z\right)^2=x^2\)
Khi đó \(A=\frac{x^2+y^2+x^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}\)
\(=1\)
Vậy \(x+y+z=0\)thì \(A=1\)
Cho mình hỏi tại sao x + y = -z \(\Rightarrow\)x-y = z
Rút gọn \(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\) biết rằng x + y + z = 0
x^2+y^2+z^2/y^2-2yx+z^2+z^2-2xy+x^2+x^2-2xy+y^2=x^2+y^2+z^2/2y^2+2x^2+2z^2-6xy=x^2+y^2+z^2/2(x^2+y^2+z^2)-6xy=1/2-6xy
xét mẫu ta có
=y^2 - 2yz + z^2 + z^2 -2xz + x^2 + x^2 -2xy +y^2
thêm bớt x^2,y^2,z^2 vào mẫu ta có
=3y^2 + 3x^2 + 3z^2 - (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)
đúng không
mà (x+y+z)=0 => (x+y+z)^2=0
mà (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz) phân tích ra thành (x+y+z)^2
=> (x^2 + y^2 + z^2 + 2xy + 2yz + 2xz)=0
=> (x^2 + y^2 + z^2 )/ 3(x^2 + y^2 + z^2)
rút gọn thành 1/3
nhớ k nha chuẩn 100%