cho x,y,z ≠0 và đôi một khác nhau thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\). . CMR: \(\left(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)=xy+yz+zx\)
Cho x,y,z khác 0 thỏa mãn x+yz=2022 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2022\)
CMR: \(\dfrac{1}{x^{2021}}+\dfrac{1}{y^{2021}}+\dfrac{1}{z^{2021}}=\dfrac{1}{x^{2021}+y^{2021}+z^{2021}}\)
Cho \(x\), \(y\), \(z\) là 3 số khác 0 thoả mãn \(x\) \(+\) \(y\) \(+\) \(z\) \(=0\). Chứng minh rằng:
\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)=\(\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
Có VT = \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{zx}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xyz}\left(x+y+z\right)}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|=VP\) (Vì x + y + z = 0)
Cho x,y,z>0 thỏa mãn x+y+z=1.CMR:\(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{9}{4}\)
từ đề bài ta có bất đẳng thức cần chứng minh tương đương:
\(3+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{9}{4}\)
<=>\(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
ta có \(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{3}{4}+\dfrac{z+y}{4x}+\dfrac{x+z}{4y}+\dfrac{x+y}{4z}=\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{4}=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(đpcm\right)\)Dấu "=" xảy ra khi x=y=z=\(\dfrac{1}{3}\)
Cho x,y,z khác 0 thỏa mãn \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
Tính P=(x+y+2z)2018
giúp mình vs ạ!!!
a. Cho x,y,z là 3 số khác 0 thỏa mãn \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
Tính giá trị biểu thức A=\(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\)
b. Cho a,b,c là các số hữu tỉ khác nhau từng đôi một. Chứng minh rằng A=\(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\)
là bình phương của 1 số hữu tỉ
c. Tìm giá trị lớn nhất của biểu thức B=\(\dfrac{5x^2+4x-1}{x^2}\)
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 3 số dương x,y,z thỏa mãn \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\)=4. Chứng minh:
\(\dfrac{1}{2x+y+z}\)+\(\dfrac{1}{x+2y+z}\)+\(\dfrac{1}{x+y+2z}\)≤1
Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)
Cộng vế với vế ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)
Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)
cho x,y,z là các số thực khác 0 thỏa mãn
\(\left\{{}\begin{matrix}\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\\x+y+z=1\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}>0\end{matrix}\right.\)
tính P=\(x^{2023}+y^{2023}+z^{2023}\)
Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)
\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)
<=> (xy)2 + (yz)2 + (zx)2 + 2xyz = (xyz)2
<=> (xy)2 + (yz)2 + (xz)2 + 2xyz(x + y + z) = (xyz)2
<=> (xy + yz + zx)2 = (xyz)2
<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)
+) Khi xy + yz + zx = -xyz
=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)
=> xy + yz + zx = xyz
<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)
<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)
<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)
<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
Khi x = -y => y = 1 => P = 1
Tương tự y = -z ; z = -x được P = 1
Vậy P = 1
