Giả sử x + y + z=2017 và \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{1}{672}\)
Tính tổng C = \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\)
Xin lỗi vì đăng không đúng dạng bài nhưng mk mong các bn giúp đỡ. Mk cảm ơn!!
Bài 1: Giả sử x + y + z = 2017 và \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{1}{672}\)
Tính tổng C = \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\)
Biết x+y+z = 2017 và \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{1}{672}\)
Tính tổng A = \(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\)
Lời giải:
\(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(A+3=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{x+y}+1\right)\)
\(A+3=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\)
\(A+3=2017\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(A+3=2017.\frac{1}{672}=\frac{2017}{672}\)
\(\Rightarrow A=\frac{2017}{672}-3=\frac{1}{672}\)
Cho x,y,z khác 0 và x+y+z khác 0. CM nếu \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\) thì \(\dfrac{1}{x^{2017}}+\dfrac{1}{y^{2017}}+\dfrac{1}{z^{2017}}=\dfrac{1}{x^{2017}+y^{2017}+z^{2017}}\)
Chào bạn
bạn nhân chéo lên rồi tách ra thì bạn sẽ có
1/x+1/y+1/z=1/x+y+z tương đương với (x+y)(y+z)(x+z)=0
Đến đây thì dễ rồi
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
cho x , y, z ≠0 thỏa mãn \(\dfrac{x+y-z}{z}\)=\(\dfrac{y+z-x}{x}\)=\(\dfrac{z+x-y}{y}\). tính P=(1+\(\dfrac{x}{y}\)).(1 +\(\dfrac{y}{z}\)).(1+\(\dfrac{z}{x}\))
Lời giải:
Nếu $x+y+z=0$ thì:
$\frac{x+y-z}{z}=\frac{-z-z}{z}=-2$
$\frac{y+z-x}{x}=\frac{-x-x}{x}=-2$
$\frac{z+x-y}{y}=\frac{-y-y}{y}=-2$
(thỏa mãn đkđb)
Khi đó:
$P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=\frac{(x+y)(y+z)(z+x)}{xyz}$
$=\frac{(-z)(-x)(-y)}{xyz}=\frac{-xyz}{xyz}=-1$
Nếu $x+y+z\neq 0$
Áp dụng TCDTSBN:
$\frac{x+y-z}{z}=\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z+y+z-x+z+x-y}{z+x+y}=\frac{x+y+z}{x+y+z}=1$
$\Rightarrow x+y=2z; y+z=2x, z+x=2y$. Khi đó:
$P=\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{2z.2x.2y}{xyz}=8$
Tìm x, y, z
\(\dfrac{x+y+2017}{z}=\dfrac{y+z-2018}{x}=\dfrac{z+x+1}{y}=\dfrac{2}{x+y+z}\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x+y+2017}{z}=\dfrac{y+z-2018}{x}=\dfrac{z+x+1}{y}=\dfrac{x+y+2017+y+z-2018+z+x+1}{z+x+y}=\dfrac{2x+2y+2z}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\dfrac{z+x+1}{y}=\dfrac{2}{x+y+z};\dfrac{z+x+1}{y}=2\\ \Rightarrow\dfrac{2}{x+y+z}=2\\ \Rightarrow x+y+z=1\)
\(\left\{{}\begin{matrix}\dfrac{x+y+2017}{z}=2\\\dfrac{y+z-2018}{x}=2\\\dfrac{z+x+1}{y}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+2017=2z\\y+z-2018=2x\\z+x+1=2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+z=3z-2017\\y+z+x=3x+2018\\z+x+y=3y-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3z-2017=1\\3x+2018=1\\3y-1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3z=2018\\3x=-2017\\3y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}z=\dfrac{2018}{3}\\x=\dfrac{-2017}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{-2017}{3}\\y=\dfrac{2}{3}\\z=\dfrac{2018}{3}\end{matrix}\right.\)
Cho các số dương x,y,z thỏa mãn: \(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{z}=\dfrac{1}{x+y-z}=\dfrac{2020}{2021}\)
Tính giá trị biểu thức \(M=\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}-\dfrac{1}{\sqrt{z}}+\dfrac{1}{\sqrt{x+y-z}}\)
Lời giải:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{x+y-z}\Leftrightarrow \frac{x+y}{xy}=\frac{1}{z}+\frac{1}{x+y-z}=\frac{x+y}{z(x+y-z)}\)
\(\Leftrightarrow (x+y)(\frac{1}{xy}-\frac{1}{z(x+y-z)})=0\)
\(\Leftrightarrow (x+y).\frac{z(x+y-z)-xy}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y).\frac{(z-x)(y-z)}{xyz(x+y-z)}=0\)
\(\Leftrightarrow (x+y)(z-x)(y-z)=0\)
Xét các TH sau:
TH1: $x+y=0$. TH này loại do ĐKXĐ $x,y>0$
TH2: $z-x=0\Leftrightarrow z=x$
$\Leftrightarrow \frac{1}{y}=\frac{2020}{2021}$
\(M=\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{y}}=2\sqrt{\frac{2020}{2021}}\)
TH3: $y-z=0$ tương tự TH2, ta có \(M=2\sqrt{\frac{2020}{2021}}\)
1) Rút gọn bt:
(x+y+z)3+(x-y-z)3+(y-x-z)3+(z-y-x)3
2)Tìm x,y,z t/m: 9x2+y2+2z2-18x+4z-6y+20=0
3)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\)=1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)=0 . CMR:
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)=1
Cho x, y, z đôi một khác nhau và \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\) = 0
Tính giá trị của biểu thức: M = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
Giúp mk giải bài này với, khó quá :((
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)