cho 1/x+1/y+1/z=0. tính giá trị biểu thức c=xy/z^2+yz/x^2+xz/y^2
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21 tháng 12 2020
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
NT
1
16 tháng 10 2023
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\left(\dfrac{1}{z}\right)^3=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}+3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)
\(\Rightarrow\dfrac{1}{z^3}=-\dfrac{1}{x^3}-\dfrac{3}{x^2y}-\dfrac{3}{xy^2}-\dfrac{1}{y^3}\)
\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot-\dfrac{1}{z}\)
\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{x^3}+\dfrac{1}{y^3}=3\cdot\dfrac{1}{xyz}\)
\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)
\(\Rightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)
\(\Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy \(A=3\)
9 tháng 9 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)
Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 3 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
22 tháng 2 2022
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
Giả thiết tương đương xy + yz + zx = 0.
Từ đó dễ dàng chứng minh được \(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3xy.yz.zx=3x^2y^2z^2\Leftrightarrow\dfrac{\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3}{3x^2y^2z^2}=\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{zx}{y^2}\).