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TG
18 tháng 7 2021
đăng ít thôi bạn! Nếu bạn đăng lẻ ra thì bn sẽ nhận đc sự trợ giúp nhanh hơn !
18 tháng 7 2021
a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
15 tháng 2 2018
a) \(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)
\(\Rightarrow|x+\frac{3}{4}|=|y-\frac{1}{5}|=|x+y+z|=0\)
\(\Rightarrow|x+\frac{3}{4}|=0\) \(\Rightarrow|y-\frac{1}{5}|=0\) \(\Rightarrow|x+y+z|=0\)
\(\Rightarrow x+\frac{3}{4}=0\) \(\Rightarrow y-\frac{1}{5}=0\) \(\Rightarrow x+y+z=0\)
\(x=\frac{-3}{4}\) \(y=\frac{1}{5}\) thay x=-3/4; y=1/5 vào biểu thức trên
ta có \(\frac{-3}{4}+\frac{1}{5}+z=0\)
\(z=0-\frac{-3}{4}-\frac{1}{5}\)
VẬY X=-3/4; Y=1/5; Z=11/20
B) \(|3x-4|+\left|3y-5\right|=0\)
\(\Rightarrow\left|3x-4\right|=\left|3y-5\right|=0\)
\(\Rightarrow\left|3x-4\right|=0\) \(\Rightarrow\left|3y-5\right|=0\)
\(3x-4=0\) \(3y-5=0\)
\(3x=4\) \(3y=5\)
\(x=\frac{4}{3}\) \(y=\frac{5}{3}\)
VẬY X= 4/3; Y=5/3
C) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)
ĐỂ \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|;\left|y-\frac{2}{5}\right|;\left|z+\frac{1}{2}\right|< 0\)
MÀ GIÁ TRỊ TUYỆT ĐỐI LUÔN MANG SỐ NGUYÊN DƯƠNG
\(\Rightarrow x;y;z\in\varnothing\)
d) \(\left|x+\frac{1}{5}\right|+\left|3-y\right|=0\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=\left|3-y\right|=0\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=0\) \(\Rightarrow\left|3-y\right|=0\)
\(x+\frac{1}{5}=0\) \(3-y=0\)
\(x=\frac{-1}{5}\) \(y=3\)
VẬY X= -1/5; Y=3
CHÚC BN HỌC TỐT!!!!!!!
15 tháng 2 2018
Ta có :
\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}\)
Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)
QS
0
MV
1 tháng 8 2017
a,
\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)
d,
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
TP
1
30 tháng 9 2017
a/ Ta có :
\(\left\{{}\begin{matrix}\left(x-1\right)^4\ge0\\\left(y-3\right)^4\ge0\end{matrix}\right.\)
Mà \(\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^4=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy ................
b/ Ta thấy :
\(\left\{{}\begin{matrix}\left(x+y\right)^{2006}\ge0\\2000\left|y-1\right|\ge0\end{matrix}\right.\)
Mà \(\left(x+y\right)^{2006}+2000\left|y-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^{2006}=0\\2000\left|y-1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\\left|y-1\right|=0\end{matrix}\right.\)
+) \(\left|y-1\right|=0\)
\(\Leftrightarrow y-1=0\)
\(\Leftrightarrow y=1\)
Mà \(x+y=0\)
\(\Leftrightarrow x=-1\)
Vậy ........
c/ Tương tự như b
TP
1
TD
30 tháng 9 2017
NX:\(\left(x-1\right)^4\ge0\forall x\)
\(\left(y-3\right)^4\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^4+\left(y-3\right)^4\ge0\forall x,y\)
\(\Rightarrow\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
b)làm tương tự phần a:
NX :|y-1| \(\ge\)0 với mọi y
=> 2000|y-1|\(\ge\)0 với mọi y
(x+y)^2006\(\ge\)0 với mọi x
=> 2000|y-1|+ (x+y)^2006\(\ge\)0 với mọi x,y
=> 2000|y-1|+ (x+y)^2006=0
<=> \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c) nhận xét |x-y-5| lớn hơn hoặc bằng 0 rồi làm tương tự
TP
1
28 tháng 5 2022
a: \(\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
b: \(\left(x+y\right)^{2006}+2000\left|y-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c: \(\left|x-y-5\right|+\left(y+3\right)^{2000}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=5\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=-3+5=2\\y=-3\end{matrix}\right.\)
\(\left(x+3\right).\left(y-4\right)=0\)
TH1:
\(x+3=0\)
=> \(x=-3\)
TH2:
\(y-4=0\)
\(y=0+4\)
\(y=4\)
Vậy x = -3; y = 4