(x-2019)2019=(x-2019)2018
Tìm x
K
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HA
4
21 tháng 7 2019
Vì \(x=2018\Rightarrow x+1=2019\)
Thay x+1=2019 vào biểu thức A ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)
\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)
\(=1\)
KN
21 tháng 7 2019
\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)
\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)
\(-2018x-x+2019\)
\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)
\(+x\left(x-2018\right)-\left(x-2018\right)+1\)
= 1
AH
Akai Haruma
Giáo viên
8 tháng 12 2021
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
KK
2
KK
14 tháng 2 2019
x2019-2019.x2018+2019.x2018+2019.x2017-2019.x2016+......2019.x-200 Tại x=2018
Giúp mik vs nhé
14 tháng 2 2019
Sai đề nên t sửa luôn nhé!
Vì \(x=2018\Rightarrow2019=2018+1=x+1\)
\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)
\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)
\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)
\(\Rightarrow A=x-200=2018-200=1818\)
NV
1
23 tháng 12 2020
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 \(\dfrac{x^{2019}+y^{2019}+z^{2019}}{a^{2019}+b^{2019}+c^{2019}}=\dfrac{x^{2019}}{a^{2019}}+\dfrac{y^{2019}}{b^{2019}}+\dfrac{z^{2019}}{c^{2019}}\)
ND
0
10 tháng 1 2021
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: \(x+y=0\).
\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).
VQ
24 tháng 4 2022
2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201
=2019*(45+54+1)-2018
=2019*100-2018
=201900-2018
=199882
24 tháng 4 2022
2019 x 45 + 54 x 2019 + 2019/2019 x 2022 - 2018 x 201
=2019*(45+54+1)-2018
=2019*100-2018
=201900-2018
=199882
28 tháng 8 2019
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
\(\left(x-2019\right)^{2019}=\left(x-2019\right)^{2018}\)
\(\Leftrightarrow\left(x-2019\right)^{2019}-\left(x-2019\right)^{2018}=0\)
\(\Leftrightarrow\left(x-2019\right)^{2018}\left(x-2019-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2019\\x=2020\end{cases}}\)
Ta có: \(\left(x-2019\right)^{2019}=\left(x-2019\right)^{2018}\)
\(\Leftrightarrow\left(x-2019\right)^{2019}-\left(x-2019\right)^{2018}=0\)
\(\Leftrightarrow\left(x-2019\right)^{2018}.\left(x-2019-1\right)=0\)
\(\Leftrightarrow\left(x-2019\right)^{2018}.\left(x-2020\right)=0\)
\(\Rightarrow\)\(x-2020=0\)Hoặc \(\left(x-2019\right)^{2018}=0\)
\( TH1:x-2020=0\Rightarrow x=2020\)
\(TH2:\left(x-2019\right)^{2018}=0\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
Vậy x= 2019 và x=2020
#Học tốt