Cho x+y+z=3 và x^2+y^2+z^2=9 tính P= (yz/x^2 +xz/y^2 +xy/z^2 -4)^2019 18/08/2021 Bởi Bella Cho x+y+z=3 và x^2+y^2+z^2=9 tính P= (yz/x^2 +xz/y^2 +xy/z^2 -4)^2019
Đáp án: $P=-1$ Giải thích các bước giải: Ta có: $2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=0\rightarrow xy+yz+zx=0$ $\rightarrow \dfrac{xy+yz+zx}{xyz}=0\rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$ Lại có: $A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}$ $\rightarrow A=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$ $\rightarrow A=xyz((\dfrac{1}{x}+\dfrac{1}{y})^3-3\dfrac{1}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3})$ $\rightarrow A=xyz((\dfrac{-1}{z})^3-3\dfrac{1}{xy}(\dfrac{-1}{z})+\dfrac{1}{z^3})$ $\rightarrow A=xyz(3\dfrac{1}{xyz})$ $\rightarrow A=3$ $\rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$ $\rightarrow (\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}-4)^{2019}=(-1)^{2019}=-1$ Bình luận
Đáp án: Ta có: 2(xy+yz+zx)=(x+y+z)2−(x2+y2+z2)=0→xy+yz+zx=02(xy+yz+zx)=(x+y+z)2−(x2+y2+z2)=0→xy+yz+zx=0 →xy+yz+zxxyz=0→1x+1y+1z=0→xy+yz+zxxyz=0→1x+1y+1z=0 Lại có: A=yzx2+xzy2+xyz2A=yzx2+xzy2+xyz2 →A=xyz(1x3+1y3+1z3)→A=xyz(1×3+1y3+1z3) →A=xyz((1x+1y)3−31xy(1x+1y)+1z3)→A=xyz((1x+1y)3−31xy(1x+1y)+1z3) →A=xyz((−1z)3−31xy(−1z)+1z3)→A=xyz((−1z)3−31xy(−1z)+1z3) →A=xyz(31xyz)→A=xyz(31xyz) →A=3→A=3 →yzx2+xzy2+xyz2=3→yzx2+xzy2+xyz2=3 →(yzx2+xzy2+xyz2−4)2019=(−1)2019=−1 Giải thích các bước giải: Bình luận
Đáp án: $P=-1$
Giải thích các bước giải:
Ta có:
$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=0\rightarrow xy+yz+zx=0$
$\rightarrow \dfrac{xy+yz+zx}{xyz}=0\rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0$
Lại có:
$A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}$
$\rightarrow A=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{1}{x}+\dfrac{1}{y})^3-3\dfrac{1}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{-1}{z})^3-3\dfrac{1}{xy}(\dfrac{-1}{z})+\dfrac{1}{z^3})$
$\rightarrow A=xyz(3\dfrac{1}{xyz})$
$\rightarrow A=3$
$\rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$
$\rightarrow (\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}-4)^{2019}=(-1)^{2019}=-1$
Đáp án:
Ta có:
2(xy+yz+zx)=(x+y+z)2−(x2+y2+z2)=0→xy+yz+zx=02(xy+yz+zx)=(x+y+z)2−(x2+y2+z2)=0→xy+yz+zx=0
→xy+yz+zxxyz=0→1x+1y+1z=0→xy+yz+zxxyz=0→1x+1y+1z=0
Lại có:
A=yzx2+xzy2+xyz2A=yzx2+xzy2+xyz2
→A=xyz(1x3+1y3+1z3)→A=xyz(1×3+1y3+1z3)
→A=xyz((1x+1y)3−31xy(1x+1y)+1z3)→A=xyz((1x+1y)3−31xy(1x+1y)+1z3)
→A=xyz((−1z)3−31xy(−1z)+1z3)→A=xyz((−1z)3−31xy(−1z)+1z3)
→A=xyz(31xyz)→A=xyz(31xyz)
→A=3→A=3
→yzx2+xzy2+xyz2=3→yzx2+xzy2+xyz2=3
→(yzx2+xzy2+xyz2−4)2019=(−1)2019=−1
Giải thích các bước giải: