\(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2016}-1}{y-2016}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
Giải pt:
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành:
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)
\(\Rightarrow x=2018;y=2019;z=2020\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)
\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)
\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)
\(x=2018,y=2019,z=2020\)
ĐK : \(\hept{\begin{cases}x>2014\\y>2015\\z>2016\end{cases}}\)
\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2014}-1}{x-2014}+\frac{1}{4}-\frac{\sqrt{y-2015}-1}{y-2015}+\frac{1}{4}-\frac{\sqrt{z-2016}-1}{z-2016}=0\)
\(\Leftrightarrow\frac{x-2010-4\sqrt{x-2014}}{4\left(x-2014\right)}+\frac{y-2011-4\sqrt{y-2015}}{4\left(y-2015\right)}+\frac{z-2012-4\sqrt{z-2016}}{4\left(x-2014\right)}=0\)
\(\Leftrightarrow\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}+\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}+\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}=0\)( 1 )
Mà \(\hept{\begin{cases}\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}\ge0\forall x>2014\\\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}\ge0\forall y>2015\\\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}\ge0\forall z>2016\end{cases}}\)( 2 )
Từ ( 1 ) và ( 2 ) => \(\hept{\begin{cases}\left(2-\sqrt{x-2014}\right)^2=0\\\left(2-\sqrt{y-2015}\right)^2=0\\\left(2-\sqrt{z-2016}\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}\sqrt{x-2014}=2\\\sqrt{y-2015}=2\\\sqrt{z-2016}=2\end{cases}}\)<=>\(\hept{\begin{cases}x=2018\\y=2019\\z=2020\end{cases}}\)( tmđk )
Vậy ( x ; y ; z ) = ( 2018 ; 2019 ; 2020 )
\(GPT\)
\(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
HELP ME
đặt x-2016=a
y-2017=b
z-2018=c
ta có\(\frac{1}{\sqrt{a}}-\frac{1}{a}+\frac{1}{\sqrt{b}}-\frac{1}{b}+\frac{1}{\sqrt{c}}-\frac{1}{c}=\frac{3}{4}\)
=>\(\left(\frac{1}{\sqrt{a}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{c}}-\frac{1}{2}\right)^2=0\)
=>\(a=b=c=4\)
còn lại tự lm nốt
Đặt \(\hept{\begin{cases}a=\sqrt{x-2009}\\b=\sqrt{y-2010}\\c=\sqrt{z-2011}\end{cases}}\)(với a,b,c>0). Khi đó phương trình đã cho trở thành
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2\)
\(\Leftrightarrow a=b=c=2\)\(\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)
Cho ba số thực dương x,y,z thỏa mãn xy+xz+yz=2016
\(\sqrt{\frac{yz}{x^2+2016}}+\sqrt{\frac{xy}{y^2+2016}}+\sqrt{\frac{xz}{z^2+2016}}\le\frac{3}{2}\)
thay 2016=xy+yz+xz vào các mẫu
dùng Cô-Si đảo vào từng phân số
sẽ dễ dàng chứng minh đc :D
Ta có
\(\sqrt{\frac{yz}{x^2+2016}}=\sqrt{\frac{yz}{x^2+yz+xy+xz}}\)
=\(\sqrt{\frac{yz}{\left(x+z\right)\left(x+y\right)}}\)\(\le\frac{1}{2}.\frac{y}{x+y}+\frac{1}{2}.\frac{z}{x+z}\)
Tương tự \(\sqrt{\frac{xy}{y^2+2016}}\le\)\(\frac{1}{2}\left(\frac{x}{y+x}+\frac{y}{y+z}\right)\)
\(\sqrt{\frac{xz}{z^2+2016}}\le\)\(\frac{1}{2}\left(\frac{x}{z+x}+\frac{z}{z+y}\right)\)
=> \(VT\)\(\le\)\(\frac{1}{2}\)(\(\frac{x}{x+y}+\frac{y}{x+y}+\frac{x}{x+z}+\frac{z}{x+z}\)+\(\frac{y}{y+z}+\frac{z}{y+z}\))
=\(\frac{3}{2}\)(\(ĐPCM\))
cho x>2016 và y>2016 thỏa mãn \(\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}\)
tính giá trị của biểu thức P=\(\frac{\sqrt{x+y}}{\sqrt{x-2016}+\sqrt{y-2016}}\)
Có :\(\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}\Rightarrow2016=\frac{xy}{x+y}\)
Do Đó :P =\(\frac{\sqrt{x+y}}{\sqrt{x-2016}+\sqrt{y-2016}}\)
\(\Leftrightarrow\)P =\(\frac{\sqrt{x+y}}{\sqrt{x-\frac{xy}{x+y}}+\sqrt{y-\frac{xy}{x+y}}}\)
\(\Leftrightarrow P=\frac{\sqrt{x+y}}{\sqrt{\frac{x^2+xy-xy}{x+y}}+\sqrt{\frac{y^2+xy-xy}{x+y}}}\)
\(\Leftrightarrow\)P =\(\frac{\sqrt{x+y}}{\sqrt{\frac{x^2}{x+y}}+\sqrt{\frac{y^2}{x+y}}}\)
\(\Leftrightarrow P=\frac{\sqrt{x+y}}{\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{x+y}}}\) (vì x;y dương )
\(\Leftrightarrow P=\frac{\sqrt{x+y}}{\frac{x+y}{\sqrt{x+y}}}\)\(\Leftrightarrow P=\frac{\sqrt{x+y}}{\sqrt{x+y}}\)
\(\Leftrightarrow P=1\)
Cho a,b,c >0; biết \(\hept{\begin{cases}a^2=b+4032\\x+y+z=a\\x^2+y^2+z^2=b\end{cases}}\)
\(P=x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}+y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{\left(2016+y^2\right)}}+z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{\left(2016+z^2\right)}}\)
Chứng minh giá trị của P không phụ thuộc vào x,y,z
Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)
Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)
\(\Rightarrow xy+yz+zx=2016\)thay vào :
\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)
\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0
Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)
\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)
Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)
Cho ba số thực dương x, y, z thỏa mãn xy+xz+yz = 2016. Chứng minh:
\(\sqrt{\frac{yz}{x^2+2016}}\)+\(\sqrt{\frac{xy}{y^2+2016}}\)+\(\sqrt{\frac{xz}{z^2+2016}}\)\(\le\)\(\frac{3}{2}\)
B1:x^2+2016=xy+yz+xz+x^2=...
tuong tu
y^2+2016=... ; z^2+2016=....
B2:bdt am-gm
(Bắc Giang)
Cho \(x,y,z\) là ba số dương thỏa mãn điều kiện \(xy+yz+zx=2016\). Chứng minh rằng
\(\sqrt{\frac{yz}{x^2+2016}}+\sqrt{\frac{zx}{y^2+2016}}+\sqrt{\frac{xy}{z^2+2016}}\le\frac{3}{2}\).
Cho x,y,x là các số thỏa mãn xyz=2016
CMR: \(\frac{2016\cdot x}{x\cdot y+2016\cdot x+2016}+\frac{y}{y\cdot z+y+2016}+\frac{z}{x\cdot z+z+1}=1\)
\(\frac{2016.x}{xy+2016x+2016}+\frac{y}{yz+y+2016}+\frac{z}{xz+z+1}\)= \(\frac{2016x}{xy+2016x+1}+\frac{xy}{xyz+xy+2016x}+\frac{xyz}{xxyz+xyz+xy}\) = \(\frac{2016x}{xy+2016x+xyz}+\frac{xy}{xyz+xy+2016x}+\frac{xyz}{2016x+xyz+xy}\)
=\(\frac{2016x+xy+xyz}{2016x+xy+xyz}=1\)
Cho \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\).Chứng minh \(\frac{x^{2016}}{a^{2016}}+\frac{y^{2016}}{b^{2016}}+\frac{z^{2016}}{c^{2016}}=\frac{x^{2016}+y^{2016}+z^{2016}}{a^{2016}+b^{2016}+c^{2016}}\)