Bài 1:

Đặt 2018=a

\(B=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)

\(=1+a-\dfrac{a}{a+1}+\dfrac{a}{a+1}=1+a=2019\)

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

\(\dfrac{x}{2017}=\dfrac{y}{2018}=\dfrac{z}{2019}=k\\ \Rightarrow\left\{{}\begin{matrix}x=2017k\\y=2018k\\z=2019k\end{matrix}\right.\)

\(4\left(x-y\right)\left(y-z\right)=4\left(2017k-2018k\right)\left(2018k-2019k\right)=4\left(-k\right)\left(-k\right)=4k^2=\left(2k\right)^2=\left(2019k-2017k\right)^2=\left(z-x\right)^2\left(ĐPCM\right)\)

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.

Lời giải:
Đặt \(\frac{x}{a}=m; \frac{y}{b}=n; \frac{z}{c}=p\). Khi đó:

ĐKĐB $\Leftrightarrow \frac{a^2m^2+b^2n^2+c^2p^2}{a^2+b^2+c^2}=m^2+n^2+p^2$

$\Rightarrow a^2m^2+b^2n^2+c^2p^2=(a^2+b^2+c^2)(m^2+n^2+p^2)$

$\Leftrightarrow a^2n^2+a^2p^2+b^2m^2+b^2p^2+c^2m^2+c^2n^2=0$
$\Rightarrow an=ap=bm=bp=cm=cn=0$

Vì $a,b,c\neq 0$ nên $m=n=p=0$

$\Rightarrow x=y=z=0$

Khi đó:

$\frac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=0$

$\frac{x^{2019}}{a^{2019}}=\frac{y^{2019}}{b^{2019}}=\frac{z^{2019}}{c^{2019}}=0$

$\Rightarrow$ đpcm

Lời giải:

PT \(\Leftrightarrow 2\sqrt{x+2018}+2\sqrt{y-2019}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow (x+2018-2\sqrt{x+2018}+1)+(y-2019-2\sqrt{y-2019}+1)+(z-2-2\sqrt{z-2}+1)=0\)

\(\Leftrightarrow (\sqrt{x+2018}-1)^2+(\sqrt{y-2019}-1)^2+(\sqrt{z-2}-1)^2=0\)

Vì \((\sqrt{x+2018}-1)^2\geq 0; (\sqrt{y-2019}-1)^2\geq 0; (\sqrt{z-2}-1)^2\geq 0\). Do đó để tổng của chúng bằng $0$ thì:

\((\sqrt{x+2018}-1)^2=(\sqrt{y-2019}-1)^2=(\sqrt{z-2}-1)^2= 0\)

\(\Rightarrow \left\{\begin{matrix} x=-2017\\ y=2020\\ z=3\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}a\ne0\\b\ne0\\c\ne0\end{matrix}\right.\)Ta có: \(\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)\cdot\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=\left(a^2+b^2+c^2\right)\cdot\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\right)\)

\(\Leftrightarrow x^2+y^2+z^2=x^2+\dfrac{x^2\cdot\left(b^2+c^2\right)}{a^2}+y^2+\dfrac{y^2\left(a^2+c^2\right)}{b^2}+z^2+\dfrac{z^2\cdot\left(a^2+b^2\right)}{c^2}\)

\(\Leftrightarrow x^2\cdot\dfrac{b^2+c^2}{a^2}+y^2\cdot\dfrac{a^2+c^2}{b^2}+z^2\cdot\dfrac{a^2+b^2}{c^2}=0\)(1)

Vì (1) luôn không âm mà a,b,c≠0

nên x=y=z=0

⇒\(\dfrac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=\dfrac{0^{2019}+0^{2019}+0^{2019}}{a^{2019}+b^{2019}+c^{2019}}=0\)

mà \(\dfrac{x^{2019}}{a^{2019}}+\dfrac{y^{2019}}{b^{2019}}+\dfrac{z^{2019}}{c^{2019}}=\dfrac{0^{2019}}{a^{2019}}+\dfrac{0^{2019}}{b^{2019}}+\dfrac{0^{2019}}{c^{2019}}=0\)

nên \(\dfrac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=\dfrac{x^{2019}}{a^{2019}}+\dfrac{y^{2019}}{b^{2019}}+\dfrac{z^{2019}}{c^{2019}}\)