Vì \(x=2018\Rightarrow x+1=2019\)

Thay x+1=2019 vào biểu thức A  ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)

\(=1\)

\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)

\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)

\(-2018x-x+2019\)

\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)

\(+x\left(x-2018\right)-\left(x-2018\right)+1\)

= 1

Vì \(x=2018\Rightarrow x+1=2019\)

Thay \(x+1=2019\) vào biểu thức \(A\) ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x-1\right)\)

\(=x^6-x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x+1\)

\(=1\)

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

x²⁰¹⁹-2019.x²⁰¹⁸+2019.x²⁰¹⁸+2019.x²⁰¹⁷-2019.x²⁰¹⁶+......2019.x-200     Tại x=2018

Giúp mik vs nhé 

Sai đề nên t sửa luôn nhé!

Vì \(x=2018\Rightarrow2019=2018+1=x+1\)

\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)

\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)

\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)

\(\Rightarrow A=x-200=2018-200=1818\)

có làm mới có ăn hỏi cc

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

Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).

Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).

Do đó dễ dàng suy ra được: \(x+y=0\).

\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019*(45+54+1)-2018

=2019*100-2018

=201900-2018

=199882

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019*(45+54+1)-2018

=2019*100-2018

=201900-2018

=199882

2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201

=2019x(45+54+1)-2018

=2019x100-2018

=201900-2018

=199882

Học Tốt nhóe ☘

Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)

\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)

Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)

=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)

=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)

= xy+xz+xy+yz+xz+yz

=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)

=4038

Vậy P=4038