cho biết \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2015\)
tính GTBT \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
Cho x, y, z khác 0 thỏa mãn x+y+z=0. Tính GTBT
P=\(\frac{x^2}{x^2-y^2-z^2}+\frac{y^2}{y^2-z^2-x^2}+\frac{z^2}{z^2-x^2-y^2}\)
Cho x,y,z > 0
Và\(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2015\)
Tính
A = \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
Cho x+y+z=7. Biết \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} = 3. Tính \frac{x^{2}}{y+z} +\frac{y^{2}}{x+z} +\frac{z^{2}}{x+y}
Tính gtbt D=\(x^{2016}+\sqrt{y}+z^{2017}\) biết\(\frac{x^2+y^2+z^2}{2+3+4}=\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}\)
Áp đụng bất đẳng thức vào
\(\left(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}\right)\ge\frac{\left(x+y+z\right)^2}{2+3+4}=\frac{x^2+y^2+z^2}{2+3+4}+\frac{2\left(xz+yz+xy\right)}{2+3+4}\)
\(\Rightarrow\hept{\begin{cases}2\left(xz+yz+xy\right)=0\\\frac{x^2}{2}=\frac{y^2}{3}=\frac{z^2}{4}\end{cases}\Rightarrow x=y=z=0}\)\(\Rightarrow D=0\)
Ta có
\(\frac{x^2+y^2+z^2}{2+3+4}=\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}\)
\(\Leftrightarrow\left(\frac{x^2}{2}-\frac{x^2}{9}\right)+\left(\frac{y^2}{3}-\frac{y^2}{9}\right)+\left(\frac{z^2}{4}-\frac{z^2}{9}\right)=0\)
\(\Leftrightarrow\frac{7x^2}{18}+\frac{2y^2}{9}+\frac{5z^2}{36}=0\)
\(\Leftrightarrow x=y=z=0\)
\(\Rightarrow D=0\)
Cho biết \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2017.\)Tính \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}.\)
\(Cho\) \(x;y;z\)là các số dương thỏa mãn \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2015\)
.Hãy tính giá trị của A=\(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
cho các số x, y, z thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2015\) tìm MAX P =\(\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\)
Tính:a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
b) Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) . Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
Cho biết: \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2009\). Tính \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
Đặt \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2009,B=\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{z^2}{x+z}\)
\(=>A-B=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{x^2}{z+x}-\frac{y^2}{x+y}-\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
\(=>2009-B=\frac{x^2-y^2}{x+y}+\frac{y^2-z^2}{y-z}+\frac{z^2-x^2}{z-x}\)
\(=>2009-B=\frac{\left(x-y\right).\left(x+y\right)}{x+y}+\frac{\left(y-z\right).\left(y+z\right)}{y+z}+\frac{\left(z-x\right).\left(z+x\right)}{z+x}\)
=>2009-B=x-y+y-x+z-x
=>2009-B=(x-x)+(y-y)+(z-z)
=>2009-B=0+0+0
=>2009-B=0
=>B=2009
Vậy \(\frac{x^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}=2009\)
thông minh đấy,mới lớp 7 mà làm được bài lớp 8