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\(\dfrac{x}{z}=\dfrac{z}{y}\Rightarrow\dfrac{x.z}{z.y}=\dfrac{x}{y}=\dfrac{x^2}{z^2}=\dfrac{z^2}{y^2}=\dfrac{x^2+z^2}{y^2+z^2}\)
đăt \(\dfrac{x}{z}=\dfrac{z}{y}=k\)
=>\(\left\{{}\begin{matrix}x=zk\\z=yk\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=yk^2\\z=yk\end{matrix}\right.\)
ta có :\(\dfrac{x}{y}=\dfrac{yk^2}{y}=k^2\left(1\right)\)
lại có \(\dfrac{x^2+z^2}{y^2+z^2}=\dfrac{y^2k^4+y^2k^2}{y^2+y^2k^2}=\dfrac{y^2k^2.\left(k^2+1\right)}{y^2.\left(1+k^2\right)}=k^2\left(2\right)\)
từ (1) và (2) => ĐPCM
b) Ta có:
\(\dfrac{19}{x+y}=\dfrac{19}{y+z}=\dfrac{19}{z+x}=\dfrac{133}{10}\)
\(\Rightarrow\dfrac{133}{7\left(x+y\right)}=\dfrac{133}{7\left(y+z\right)}=\dfrac{133}{7\left(z+x\right)}=\dfrac{133}{10}\)
\(\Rightarrow7\left(x+y\right)=7\left(y+z\right)=7\left(z+x\right)=10\)
\(\Rightarrow7\left(x+y\right)+7\left(y+z\right)+7\left(z+x\right)=10\)
\(\Rightarrow7\left[2\left(x+y+z\right)\right]=10\)
\(\Rightarrow14\left(x+y+z\right)=10\)
\(\Leftrightarrow x+y+z=\dfrac{5}{7}\)
Lời giải:
\(y^2=xz\Rightarrow \frac{y}{z}=\frac{x}{y}\)
\(z^2=yt\Rightarrow \frac{z}{t}=\frac{y}{z}\)
Vậy \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}\)
Ta có:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\Rightarrow \frac{x^3}{y^3}=\frac{y^3}{z^3}=\frac{z^3}{t^3}=\frac{x^3+y^3+z^3}{y^3+z^3+t^3}(1)\) (áp dụng tính chất dãy tỉ số bằng nhau)
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\Rightarrow \frac{x^3}{y^3}=\frac{x}{y}.\frac{y}{z}.\frac{z}{t}=\frac{x}{t}(2)\)
Từ \((1);(2)\Rightarrow \frac{x^3+y^3+z^3}{y^3+z^3+t^3}=\frac{x}{t}\) (đpcm)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
\(\dfrac{x-y}{x+y}=\dfrac{z-x}{z+x}\\ \Rightarrow\left(x-y\right)\left(z+x\right)=\left(x+y\right)\left(z-x\right)\\ \Rightarrow xz+x^2-yz-yx=xz-x^2+yz-yx\\ \Rightarrow xz-xz+x^2+x^2=yz+yz-yx+yx\\ \Rightarrow2x^2=2yz\\ \Rightarrow x^2=yz\)