Bài 1:tìm x;y
a)|x-y-2|+|y+3|=0
b)|x-2007|+|y-2008|=0
c)|2/3-1/2+3/4x|+|1,5-11/17+23/13y|=0
d)|x-y-5|+|y-2| nhỏ hơn bằng 0
e)|3x+2y|+|4y-1| nhỏ hơn bằng 0
làm câu nào cg đc
K
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18 tháng 8 2021
Bài 1 :
\(\left|2x-1\right|=x-1\)ĐK : \(x\ge1\)
TH1 : \(2x-1=x-1\Leftrightarrow x=0\)(ktm)
TH2 : \(2x-1=1-x\Leftrightarrow3x=2\Leftrightarrow x=-\frac{2}{3}\)(ktm)
Vậy biểu thức ko có x thỏa mãn
Bài 2 :
\(\left|3x-1\right|=2x+3\)ĐK : x >= -3/2
TH1 : \(3x-1=2x+3\Leftrightarrow x=4\)
TH2 : \(3x-1=-2x-3\Leftrightarrow5x=-2\Leftrightarrow x=-\frac{2}{5}\)
biết:
b./ 
c*./ 
b/ 
c/ 
= 
e/ 
= 
và x + y = 22 f/ 
và 
e) 9x =81 
h) 
và x + y= -21 i) 
và 3x - 2y = -2
và 3x – 2y – z = -29
KY
1 tháng 7 2021
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
\(\left|x-y-2\right|+\left|y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|x-y-2\right|\ge0\forall x;y\\\left|y+3\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-2\right|+\left|y+3\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-y-2\right|=0\Rightarrow x-\left(-3\right)-2=0\Rightarrow x+1=0\Rightarrow x=-1\\\left|y+3\right|=0\Rightarrow y+3=0\Rightarrow y=-3\end{matrix}\right.\)
\(\left|x-2007\right|+\left|y-2008\right|=0\)
\(\left\{{}\begin{matrix}\left|x-2007\right|\ge0\forall x\\\left|y-2008\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-2007\right|+\left|y-2008\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-2007\right|=0\Rightarrow x-2007=0\Rightarrow x=2007\\\left|y-2008\right|=0\Rightarrow y-2008=0\Rightarrow y=2008\end{matrix}\right.\)
\(\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|=0\)
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|\ge0\forall x\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}y\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|+\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{3}{4}x\right|=0\Rightarrow\dfrac{1}{6}+\dfrac{3}{4}x=0\Rightarrow\dfrac{3}{4}x=-\dfrac{1}{6}\Rightarrow x=-\dfrac{2}{9}\\\left|1,5-\dfrac{11}{17}+\dfrac{23}{13}x\right|=0\Rightarrow\dfrac{29}{34}+\dfrac{23}{13}x=0\Rightarrow\dfrac{23}{13}x=-\dfrac{29}{34}\Rightarrow x=-\dfrac{377}{782}\end{matrix}\right.\)
\(\left|x-y-5\right|+\left|y-2\right|\le0\)
\(\left\{{}\begin{matrix}\left|x-y-5\right|\ge0\forall x;y\\\left|y-2\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|+\left|y-2\right|\le0\\\left|x-y-5\right|+\left|y-2\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-y-5\right|+\left|y-2\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\Rightarrow x-2-5=0\Rightarrow x=7\\\left|y-2=0\right|\Rightarrow y=2\end{matrix}\right.\)
\(\left|3x+2y\right|+\left|4y-1\right|\le0\)
\(\left\{{}\begin{matrix}\left|3x+2y\right|\ge0\forall x;y\\ \left|4y-1\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|\ge0\)
Lúc này ta có:
\(\left\{{}\begin{matrix}\left|3x+2y\right|+\left|4y-1\right|\ge0\\\left|3x+2y\right|+\left|4y-1\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|3x+2y\right|+\left|4y-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x+2y\right|=0\Rightarrow3x+\dfrac{1}{2}=0\Rightarrow3x=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{6}\\\left|4y-1\right|=0\Rightarrow4y=1\Rightarrow y=\dfrac{1}{4}\end{matrix}\right.\)