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Sai đề kìa . Đề đúng đây :
\(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}\)
Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\left(k>0\right)\)
Ta có :
x = 1998k ; y = 1999k ; z =2000k
Ta có :
\(\left(x-z\right)^3=\left(1998k-2000k\right)^3=\left(-2k\right)^3=-8k\) (*)
\(8\left(x-y\right)^2\cdot\left(y-z\right)=8\left(1998k-1999k\right)^2\cdot\left(1999k-2000k\right)\)
\(=8\left(-1\right)^2\cdot\left(-1\right)=-8\) (**)
Từ (*) và (**) suy ra ĐPCM
\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
Lời giải:
Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$
$\Rightarrow x=2018a; y=2019a; z=2020a$
$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$
Mặt khác:
$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$
Từ $(1); (2)$ ta có đpcm.
\(\frac{x}{1998}=\frac{y}{1999}=\frac{z}{2000}=t=\frac{x-z}{1998-2000}=\frac{x-y}{1998-1999}=\frac{y-z}{1999-2000}.\)
Hay: \(\frac{x-z}{-2}=\frac{x-y}{-1}=\frac{y-z}{-1}\Rightarrow x-z=2\left(x-y\right)=2\left(y-z\right)\)(1)
a) \(\left(x-z\right)^3=\left(x-z\right)^2\left(x-z\right)=\left(2\left(x-y\right)\right)^2\left(2\left(y-z\right)\right)\)
\(\Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\)ĐPCM a)
b) Từ (1) => x + z = 2y
Để \(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\Rightarrow\frac{x+y}{\frac{1}{2}}=\frac{y+z}{\frac{1}{5}}=\frac{z+x}{\frac{1}{3}}\)
Từ \(\Rightarrow\frac{x+y}{\frac{1}{2}}=\frac{y+z}{\frac{1}{5}}=\frac{x+y+y+z}{\frac{1}{2}+\frac{1}{5}}=\frac{4y}{\frac{7}{10}}=\frac{2y}{\frac{1}{3}}\)
=>y=0 =>x=0 => z=0 Suy ra hệ thức: x-y/4=y-z/5 luôn đúng. ĐPCM
ĐẶT\(\frac{x}{1998}=\frac{y}{1999}=\frac{z}{2000}=k\Rightarrow x=1998k,y=1999k,z=2000k\)
\(\Rightarrow\left(x-z\right)^3=\left(1998k-2000k\right)^3=\left(-2k\right)^3=-8k^3\)
\(8.\left(x-y\right)^2.\left(y-z\right)=8.\left(1998k-1999k\right)^2.\left(1999k-2000k\right)=-8k^3\)
=> đpcm
Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\)
\(\Rightarrow x=1998k;y=1999k;z=2000k\)
\(\left(x-z\right)^3=\left(2000k-1998k\right)^3=8k^3\)
\(8\left(x-y\right)^2\left(y-z\right)=8\left(1999k-1998k\right)^2.\left(1999k-2000k\right)\\ =8.k^2.k=8k^3\\ \Rightarrowđpcm\)