Giải chi tiết nha
Thu gọn biểu thức sau:
\(2xy-2yz\cdot z+xy+\dfrac{1}{2}z^2y+2zy\cdot z\)
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VN
1
TD
18 tháng 2 2022
\(=3xy-2yz^2+\dfrac{1}{2}yz^2+2yz^2\)
=\(3xy+\dfrac{1}{2}yz^2\)
11 tháng 2 2022
\(A=\dfrac{-1}{5}x^3\cdot\dfrac{1}{32}x^{20}y^5\cdot\dfrac{64}{27}x^3y^9\cdot z^{2022}=-\dfrac{2}{135}x^{26}y^{14}z^{2022}\)
11 tháng 2 2022
`A=\frac{-1}{5}x^3 \times \frac{1}{32}x^{20}y^5 \times \frac{64}{27}x^3y^9 \times z^{2022}=-\frac{2}{135}x^{26}y^{14}z^{2022}`
19 tháng 12 2020
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
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: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
N
9 tháng 4 2017
Áp dụng BĐT cauchy ta có:\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\x^2+z^2\ge2xz\end{matrix}\right.\)
\(P\le\dfrac{1}{4xy+4x+4}+\dfrac{1}{4yz+4y+4}+\dfrac{1}{4xz+4z+4}=\dfrac{1}{4}\left(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+x+1}\right)\)
xét biểu thức \(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{zx+z+1}=\dfrac{1}{xy+x+1}+\dfrac{x}{1+yx+x}+\dfrac{xy}{x+1+xy}=\dfrac{xy+x+1}{xy+x+1}=1\)do đó \(P\le\dfrac{1}{4}\)
dấu = xảy ra khi x=y=z=1
HN
10 tháng 4 2017
Trước tiên ta tính:
\(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\)
Đặt: \(\left\{{}\begin{matrix}x=\dfrac{a}{b}\\y=\dfrac{b}{c}\\z=\dfrac{c}{a}\end{matrix}\right.\left(a,b,c\ne0\right)\)
Thì ta có: \(\dfrac{1}{\dfrac{a}{b}+\dfrac{a}{b}.\dfrac{b}{c}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{b}{c}.\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{c}{a}.\dfrac{a}{b}+1}\)
\(=\dfrac{bc}{ab+ac+bc}+\dfrac{ca}{ab+bc+ca}+\dfrac{ab}{ab+bc+ca}=1\)
Quay về bài toán ban đầu. Ta có:
\(P=\dfrac{1}{\left(x+2\right)^2+y^2+2xy}+\dfrac{1}{\left(y+2\right)^2+z^2+2yz}+\dfrac{1}{\left(z+2\right)^2+x^2+2xz}\)
\(=\dfrac{1}{x^2+4x+4+y^2+2xy}+\dfrac{1}{y^2+4y+4+z^2+2yz}+\dfrac{1}{z^2+4z+4+z^2+2xz}\)
\(=\dfrac{1}{\left(x-y\right)^2+4x+4xy+4}+\dfrac{1}{\left(y-z\right)^2+4y+4yz+4}+\dfrac{1}{\left(z-x\right)^2+4z+4zx+4}\)
\(\le\dfrac{1}{4x+4xy+4}+\dfrac{1}{4y+4yz+4}+\dfrac{1}{4z+4zx+4}\)
\(=\dfrac{1}{4}.\left(\dfrac{1}{x+xy+1}+\dfrac{1}{y+yz+1}+\dfrac{1}{z+zx+1}\right)=\dfrac{1}{4}\)
\(=2xy-2yz^2+xy+\frac{1}{2}yz^2+2yz^2=3xy+\frac{1}{2}yz^2\)