1.tìm x,y,z (6*x + 1)2020 + (3*y + 1)2020 + |x + y +2z2 | =0
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B
1
10 tháng 7 2019
Ta có: \(\left(x-\frac{1}{5}\right)^{2020}\ge0\forall x\)
\(\left(y+0.4\right)^{2000}\ge0\forall y\)
\(\left(z-3\right)^6\ge0\forall z\)
=> \(\left(x-\frac{1}{5}\right)^{2020}+\left(y+0.4\right)^{2000}+\left(z-3\right)^6\ge0\forall x,y,z\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-\frac{1}{5}=0\\y+0.4=0\\z-3=0\end{cases}}\) => \(\hept{\begin{cases}x=\frac{1}{5}\\y=0\\z=3\end{cases}}\)
vậy ...
UI
1
7 tháng 8 2020
\(x^3+y^3+z^3+6=3\left(x^2+y^2+z^2\right)\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz+6=3\left(x^2+y^2+z^2\right)\)Mà x+y+z=3
\(\Rightarrow3\left(x^2+y^2+z^2-xy-xz-yz\right)+3xyz+6=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow x^2+y^2+z^2-xy-yz-xz+xyz+2=x^2+y^2+z^2\)
\(\Rightarrow xyz-xy-yz-xz+2=0\Rightarrow\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(2-x-y\right)=0\)
\(\Rightarrow xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(2-3+z\right)=0\Rightarrow xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)=0\)
\(\Rightarrow\left(z-1\right)\left(xy-x-y+1\right)=0\Rightarrow\left(z-1\right)\left[\left(xy-x\right)-\left(y-1\right)\right]=0\Rightarrow\left(z-1\right)\left[x\left(y-1\right)-\left(y-1\right)\right]=0\)
\(\Rightarrow\left(z-1\right)\left(x-1\right)\left(y-1\right)=0\)
Suy ra có ít nhất 1 trong 3 số x,y,z bằng 1,khi đó A=0
Vậy A=0
UI
0
9 tháng 5 2021
ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)
Đặt \(\sqrt{x-2019}=a,......\)
Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)
\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)
- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)
\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)
- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)
- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )
Vậy ...
6 tháng 10 2021
\(1,\Rightarrow2^b\left(2^{a-b}-1\right)=256=2^8\left(a>b\right)\)
Do \(2^b\) chẵn, \(2^{a-b}-1\) lẻ, \(2^8\) chẵn nên \(2^{a-b}-1=1\Leftrightarrow2^{a-b}=2\Leftrightarrow a-b=1\)
\(\Leftrightarrow2^b\cdot1=2^8\Leftrightarrow b=8\Leftrightarrow a=9\)
Vậy \(\left(a;b\right)=\left(8;9\right)\)
KN
8 tháng 10 2020
x2 + 2y2 + z2 - 2xy - 2y - 4z + 5 = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + ( z2 - 4z + 4 ) = 0
<=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2 = 0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )
Thay ( 1 ) vào A , ta được :
\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)
Vậy A = 2
8 tháng 10 2020
Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)
TT
tìm x y z thoả mãn đẳng thức 1/x2022+1/y2022+1/z2022=1/x2021+1/y2021+1/z2021=1/x2020+1/y2020+1/z2020
0
PA
2
L
18 tháng 2 2020
( x - 1 )2018 + ( y + 3 )2020 + ( z - 5 )2022 = 0
Ta thấy : ( x - 1 )2018 \(\ge0\) ; ( y + 3 )2020 \(\ge0\) ; ( z - 5 )2022 \(\ge0\)
\(\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left(z-5\right)^{2022}\ge0\)
Theo đề,ta có : \(\left(x-1\right)^{2018}=\left(y+3\right)^{2020}=\left(z-5\right)^{2022}=0\)
+) \(\left(x-1\right)^{2018}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(y+3\right)^{2020}=0\Rightarrow y+3=0\Rightarrow y=-3\)
=) \(\left(z-5\right)^{2022}=0\Rightarrow z-5=0\Rightarrow z=5\)
Vậy : x = 1 ; y = -3 ; z = 5
S
18 tháng 2 2020
\(\text{Ta có:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\\\left(y+3\right)^{2020}\ge0\\\left(z-5\right)^{2022}\ge0\end{cases}}\text{mà:}\left(x-1\right)^{2018}+\left(y-2\right)^{2020}+\left(z-3\right)^{2022}=0\text{ nên:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}=0\\\left(y+3\right)^{2018}=0\\\left(z-5\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-3\\z=5\end{cases}}\)
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