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Dễ dàng chứng minh được : nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
Ta có \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)( Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\))
Đặt bài toán phụ : Chứng minh nếu \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
Thật vậy :
\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^3=0\)
\(a+b=-c\)
\(b+c=-a\)
\(c+a=-b\)
\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=-3\left(-c\right)\left(-b\right)\left(-a\right)\)
\(=3abc\)
Trở lại bài toán chính :
Ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Rightarrow xy+xz+yz=0\)
\(\Rightarrow\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3\left(xy\right)\left(xz\right)\left(yz\right)=3x^2y^2z^2\)
Lại có:
\(P=\frac{xy.y^2x^2}{x^2y^2z^2}+\frac{xz.z^2.x^2}{x^2y^2z^2}+\frac{z^2.y^2.yz}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3}{x^2y^2z^2}+\frac{\left(xz\right)^3}{x^2y^2z^2}+\frac{\left(yz\right)^3}{x^2y^2z^2}\)
\(=\frac{\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)}{x^2y^2z^2}\)
Thay \(\left(xy\right)^3+\left(xz\right)^3+\left(yz^3\right)=3x^2y^2z^2;\)ta có:
\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}\)
\(=3\)
Vậy \(P=3.\)
Vì 1/x + 1/y + 1/z = 0 nên lần lượt nhân vs x; y; z ta có:
1 + x/y + x/z = 0 (1)
1 + y/z + y/x = 0 (2)
1 + z/x + z/y = 0 (3)
Từ (1); (2); (3) suy ra : x/y + y/z + z/x + x/z + y/x + z/y = - 3 (*)
Mặt khác : 1/x + 1/y + 1/z = 0 nên quy đồng lên ta có:
(xy + yz + zx)/xyz = 0 hay xy + yz + zx = 0
Hay : (1/x^2 + 1/y^2 + 1/z^2).(xy + yz + zx) = 0
khai triển ra :
yz/x^2 + zx/y^2 + xy/z^2 + x/y + y/z + z/x + x/z + y/x + z/y = 0
Vậy : yz/x^2 + zx/y^2 + xy/z^2 = - (x/y + y/z + z/x + x/z + y/x + z/y) = 3 (theo (*))
Đầu tiên cần chứng minh khẳng định sau : Nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\)
Thật vậy : Xét \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Áp dụng khẳng định trên với \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\)được
\(P=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)
Chú ý : Đề bài cần thêm điều kiện x,y,z khác 0
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
CMTT:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A=\(\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
\(A=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà \(xy+yz+xz=0\)
Từ \(\Rightarrow\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy A=0
a, Chứng minh \(x^3+y^3+z^3=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3\)
Biến đổi vế phải thì ta phải suy ra điều phải chứng minh
b, Ta có: \(a+b+c=0\)thì
\(a^3+b^3+c^3==\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
( Vì \(a+b+c=0\)nên \(a+b=-c\))
Theo giả thuyết \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
\(=xyz.\frac{3}{xyz}=3\)
\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)
\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)
Ta có :\(yz+xz+xy=0\)
\(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)
\(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)
\(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)
\(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{-1}{z^3}\)
\(\frac{1}{x^3}+\frac{3}{x^2y}+\frac{3}{xy^2}+\frac{1}{y^3}=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{3}{xyz}\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{3}{xyz}.xyz\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xy}{z^2}+\frac{xz}{y^2}=3\)
khi gấp lên mấy lần thì nó vẫn bằng 0 nên biểu thức đó bằng 0
trả lời thì ghi hẳn lời giải ra =)))
A= \(\frac{yz}{x^2}\)+\(\frac{xy}{z^2}\)+\(\frac{xz}{y^2}\)= xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\))
Đặt \(\frac{1}{x}\)=a ; \(\frac{1}{y}\)=b; \(\frac{1}{z}\) =c
=> a+b+c=0
Ta có a3+b3+c3- 3abc= (a+b)3+c3-3ab(a+b)-3abc
=(a+b+c)3- 3ab(a+b)-3abc-3(a+b)c(a+b+c)
=(a+b+c)3- 3ab(a+b+c)- 3(a+b)c(a+b+c)
Mà a+b+c=0 => a3+b3+c3- 3abc=0
=>a3+b3+c3=3abc
=> \(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\)=\(\frac{3}{xyz}\)
=> A=xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\))= xyz.\(\frac{3}{xyz}\)=3
(đúng thì tk nha)
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}x1+y1+z1=0⇒x1+y1=−z1
1/x + 1/y + 1/z = 0 => 1/x + 1/y = - 1/z
<=> ( 1/x + 1/y )^3 = -1/z^3
1/x^3 + 3/x^2y + 3/xy^2 + 1/y^3 = -1/z^3
<=> 1/x^3 + 1/y^3 + 1/z^3 = -3/x^2y - 3/xy^2 = -3/xy(1/x + 1/y) = 3/xyz
=> xyz = (1/x^3 + 1/y^3 + 1/z^3) =3/xyz . xyz
=> yz/x^2 + xy/z^2 + xz/y^2 = 3
x 1 + y 1 + z 1 = 0⇒ x 1 + y 1 = − z 1 ⇔ x 1 + y 1 3 = z 3 −1 x 3 1 + x 2 y 3 + xy 2 3 + y 3 1 = − z 3 1 ⇔ x 3 1 + y 3 1 + z 3 1 = x 2 y −3 − xy 2 3 = xy −3 x 1 + y 1 = xyz 3 ⇒xyz x 3 1 + y 3 1 + z 3 1 = xyz 3 .xyz ⇒ x 2 yz + z 2 xy + y 2 xz = 3
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+xz}{xzy}=0\Leftrightarrow xy+yz+xz=0\)
\(A=\frac{yz}{x^2}+\frac{xy}{z^2}+\frac{xz}{y^2}=\frac{x^3y^3+y^3z^3+x^3z^3}{x^2y^2z^2}\)
\(=\frac{x^3y^3+y^3z^3+x^3z^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}\)
\(=\frac{\left(xy+yz+xz\right)\left(x^2y^2+x^2z^2+y^2z^2-xyz\left(x+y+z\right)\right)+3x^2y^2z^2}{x^2y^2z^2}\)
\(=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
vậy \(A=3\)
CÁCH KHÁC:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Leftrightarrow yz+xz+xy=0\left(1\right)\)
\(A=\frac{yz}{^{x^2}}+\frac{xz}{y^2}+\frac{xz}{z^2}=\frac{\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3}{\left(xyz\right)^2}\)
Mặt khác \(\left(1\right)\Leftrightarrow yz+xz=-xy\)
\(\Leftrightarrow\left(yz+xz\right)^3=-xy^3\)
\(\Leftrightarrow\left(yz\right)^3+3xy^2z^3+3x^2yz^3+\left(xz\right)^2+\left(xy\right)^3=0\)
\(\Leftrightarrow\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3=-3xyz^2\left(yz+xz\right)\left(2\right)\)
Vì \(yz+xz=-xy\)nên \(\left(2\right)\Leftrightarrow\left(yz\right)^3+\left(xz\right)^3+\left(xy\right)^3=3(xyz)^2\)
Thay vào A ta được:
\(A=\frac{3\left(xyz\right)^2}{\left(xyz\right)^2}=3\)
??????????????????????????????????????
Ta có :
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) \(\Rightarrow\) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{z}+\frac{1}{x}=-\frac{1}{y}\end{cases}}\)
lại có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x}+\frac{1}{z}\right)=0\)
hay \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( 1 )
\(A=\frac{yz}{x^2}+\frac{xy}{y^2}+\frac{xz}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\) ( 2 )
Từ (1), (2) \(\Rightarrow A=3\)
đáp án bằng : 3
Cứ động não lên là làm đc đó bạn
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