Cho ba số x,y,z khác 0 thỏa mãn x+y+z=0.Tính \(\dfrac{2x\left(x+y\right)\left(z+x\right)+y\left(x+y\right)\left(y+z\right)}{z\left(x+z\right)\left(y+z\right)}\)
cho x,y,z>0 thỏa mãn \(\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=8\)
Tìm giá trị nhỏ nhất của S=\(xyz\left(x+y+z\right)^3\)
(có thể dùng BDT \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\))
tks mn<3
Cho 3 số x,y,z khác 0 thoả mãn điều kiện \(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
Hãy tính giá trị của biểu thức :
\(B=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1
--> y + z - x = x; z + x - y = y; x + y - z = z
--> y + z = 2x; z + x = 2y; x + y = 2z
Ta có:
B = (x + y)/y.(y + z)/z.(z + x)/x
= 2z/y.2x/z.2y/x = 8
cho x,y,z là các số thực khác, thỏa mãn:
\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
tính gtbt: \(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)
<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)
đặt x+y+z = t
=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)
=> x=y=z
ta lại có
\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
vì x=y=z => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Cho x, y, z đôi một khác nhau thỏa mãn \(\left(x+z\right)\left(y+z\right)=1\). Tìm Min
\(M=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(x+z\right)^2}+\dfrac{1}{\left(y+z\right)^2}\)
Cho 3 số thực dương \(x,y,z\) thỏa mãn \(x+y+z=3\). Tìm GTLN của biểu thức \(P=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)
\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)
\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)
Tương tự ta được
\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)
\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :
\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)
\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)
\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)
Cho ba số x,y,z thỏa mãn: \(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{z}{2020}\)
CMR: \(\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\)
\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
Cho x, y, z là các số thực khác 0 thỏa mãn:
\(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{x}+\dfrac{1}{z}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) và x3 + y3 + z3 =1
Tính giá trị của biểu thức P= \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
cho các số x,y,z thoả mãn \(\dfrac{x}{y-z}+\dfrac{y}{z-x}+\dfrac{z}{x-y}=0\)
tính giá trị biểu thức A=\(\dfrac{x}{\left(y-z\right)^2}+\dfrac{y}{\left(z-x\right)^2}+\dfrac{z}{\left(x-y\right)^2}\)
Lời giải:
\(A=\left(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}\right)\left(\frac{1}{y-z}+\frac{1}{z-x}+\frac{1}{x-y}\right)-\frac{x}{(y-z)(z-x)}-\frac{x}{(y-z)(x-y)}-\frac{y}{(z-x)(x-y)}-\frac{y}{(z-x)(y-z)}-\frac{z}{(x-y)(y-z)}-\frac{z}{(x-y)(z-x)}\)
\(=0-\frac{x(x-y)+x(z-x)+y(y-z)+y(x-y)+z(z-x)+z(y-z)}{(x-y)(y-z)(z-x)}\)
\(=0-\frac{x^2+xz+y^2+xy+z^2+zy-(xy+x^2+yz+y^2+zx+z^2)}{(x-y)(y-z)(z-x)}=0-\frac{0}{(x-y)(y-z)(z-x)}=0\)
Cho x,y,z >0 thỏa x+y+z=\(\sqrt{2021}\)
Tìm Min:
\(P=\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.\left(\dfrac{\sqrt{y+z}}{x}+\dfrac{\sqrt{z+x}}{y}+\dfrac{\sqrt{x+y}}{z}\right)\)
Thử nhé
Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)
Thay vo P ta duoc \(P=4.\sqrt{2021}\)
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\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)
Cauchy-Schwarz:
\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)
\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)
\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)
Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)
\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)