1. Tính tổng: S = x+ 2y+3z, biết rằng:
\(\dfrac{1}{x+2y}+\dfrac{1}{2y+3z}+\dfrac{1}{3z+x}=\dfrac{12x}{2y+3z}+\dfrac{24y}{3z+x}+\dfrac{36z}{x+2y}=2016\)
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Tính tổng \(S=x+2y+3z\), biết rằng:
\(\dfrac{1}{x+2y}+\dfrac{1}{2y+3z}+\dfrac{1}{3z+x}=\dfrac{12x}{2y+3z}+\dfrac{24y}{3z+x}-\dfrac{36z}{x+2y}=2016\)
tính tổng S = x + 2y + 3z biết rằng 1/(x+ 2y) + 1/(2y+3z)+1/(x+3z)= 12x/(2y+3z)+24y/(x+3z)+ 36z/(x+2y)=2016
Cho ba số thực dương x,y,z thoả mãn :x+2y+3z=18 .Chứng minh rằng :
\(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
Tinh tong : S= x+2y +3z, biet rang : \(\frac{1}{x+2y}+\frac{1}{2y+3z}+\frac{1}{3z+z}=\frac{12x}{2y+3z}+\frac{24y}{3z+x}-\frac{36z}{x+2y}=2016\)
Cho ba số thực x,y,z thoả mãn : x+2y+3z=18
Cmr : \(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
\(VT=\dfrac{2y+3z+5}{1+x}+1+\dfrac{3z+x+5}{2y+1}+1+\dfrac{x+2y+5}{1+3z}+1-3\)
\(VT=\dfrac{x+2y+3z+6}{1+x}+\dfrac{x+2y+3z+6}{1+2y}+\dfrac{x+2y+3z+6}{1+3z}-3\)
\(VT=24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)-3\ge\dfrac{24.9}{1+x+1+2y+1+3z}-3=\dfrac{216}{21}-3=\dfrac{51}{7}\)
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tính tổng S= x+2y+3z biết rằng:
\(\frac{1}{x+2y}\)+\(\frac{1}{2y+3z}\)+\(\frac{1}{3z+x}\)= \(\frac{12x}{2y+3z}\)+\(\frac{24y}{3z+x}\)+\(\frac{36z}{x+2y}\)=2016
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cho các số thực dương x,y,z thỏa x+2y+3z=18 CMR
\(\dfrac{2y+3z+5}{1+x}+\dfrac{3z+x+5}{1+2y}+\dfrac{x+2y+5}{1+3z}\ge\dfrac{51}{7}\)
\(VT+3=\left(x+2y+3z+6\right)\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
= \(24\left(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\right)\)
Áp dụng BĐT cauchy-schwarz:
\(\dfrac{1}{1+x}+\dfrac{1}{1+2y}+\dfrac{1}{1+3z}\ge\dfrac{9}{3+x+2y+3z}=\dfrac{9}{21}\)
\(\Rightarrow VT\ge\dfrac{24.9}{21}-3=\dfrac{51}{7}\)
dấu = xảy ra khi x=2y=3z=6 hay x=6,y=3,z=2
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cho 2 đa thức A= \(-4x^5y^3+x^4y^3-3x^2y^3z^2-x^4y^3+x^2y^3z^2-2y^4\)
a) thu gọn rồi tìm bậc đa thức A
b) tìm đa thức B biết rằng B\(-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=A\)
a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)
b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)
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26 tháng 12 2021 lúc 10:34
Cho \(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}\) và \(x-2y+3z=8\)
Tìm x, y, z
\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{2}=\dfrac{2z-4x}{3}=\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x-2y+3z}{2-6+12}=\dfrac{8}{8}=1\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=4\end{matrix}\right.\)
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