Câu 1: tìm x biết:
a) \(x^2-2y^2=1\)
b) \(|x-2018y|+\left(y-1\right)^{2018}=0\)
c) \(|x+5|+\left(3y-4\right)^{2018}=0\)
HELP ME! chiều pải nộp r
K
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LN
2
25 tháng 1 2022
Vì \(\left(x-5\right)^{2018}\ge0;\left|2y^2-162\right|^{2018}\ge0\Rightarrow\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}\ge0\)
mà \(\left(x-5\right)^{2018}+\left|2y^2-162\right|^{2018}=0\)
Dấu ''='' xảy ra khi x = 5 ; \(2y^2=162\Leftrightarrow y^2=81\Leftrightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
LH
25 tháng 1 2022
Vì \(\left(x-5\right)^{2018}\ge0\\ \left|2y^2-162\right|^{2018}\ge0\\ \)
Suy ra phương trình dc thỏa mãn khi và chỉ khi x-5 = 0 và 2y^2-162=0
\(\left\{{}\begin{matrix}\left(x-5\right)^{2018}=0\\\left|2y^2-162\right|^{2018}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\2\left(y^2-81\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\pm9\end{matrix}\right.\)
29 tháng 12 2018
a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)
\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)
\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)
Vậy........
9 tháng 7 2017
2. CMR:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)
=> đpcm.
c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)
=> đpcm.
4 tháng 1 2021
Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall y\)
\(2\left(x+y\right)^2\ge0\forall x,y\)
Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)
Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được:
\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)
\(=0^{2016}+1^{2017}+0^{2018}=1\)
Vậy: M=1
VM
1
AV
7 tháng 5 2018
Vì : (3x+1)2018+(2y-1)2018+\(\left|x+2y-z\right|\)2018=0
Nên: \(\left\{{}\begin{matrix}\left(3x+1\right)^{2018}=0\\\left(2y-1\right)^{2018}\\\left|x+2y-z\right|^{2018}=0\end{matrix}\right.=0\) ⇔\(\left\{{}\begin{matrix}3x+1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\\dfrac{-1}{3}+1-z=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{matrix}\right.\)
Vậy : x=\(\dfrac{-1}{3}\) ; y=\(\dfrac{1}{2}\) ; z=\(\dfrac{2}{3}\)
N
25 tháng 11 2018
\(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4\ge0\\\left(3x-2y\right)^2\ge0\end{cases}}\Rightarrow\left|x^2+y^2+z^2-1\right|+\left(3y-4z\right)^4+\left(3x-2y\right)^2\ge0\)
dấu = xảy ra khi \(\hept{\begin{cases}\left|x^2+y^2+z^2-1\right|=0\\\left(3y-4z\right)^4=0\\\left(3x-2y\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\3y=4z\\3x-2y=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2+y^2+z^2=1\\y=\frac{4z}{3}\\x=\frac{2y}{3}\end{cases}}\)
Vậy ...
p/s bài này chắc chỉ có dạng chung thôi bn :)
b) \(\left|x-2018y\right|+\left(y-1\right)^{2018}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-2018y\right|=0\\\left(y-1\right)^{2018}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018y=0\\y-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018y=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018.1=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2018=0\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2018\\y=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=2018\\y=1\end{matrix}\right.\)
c) \(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2018}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\3y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
giúp mk lun con d) nha:
d) (2x-1)^2 +\(|2y-x|-8=12-5.2^2\)