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a)
\(\left\{{}\begin{matrix}\left|x+5\right|\le4\\x\in Z\end{matrix}\right.\) \(\Leftrightarrow-4\le x+5\le4\Rightarrow-9\le x\le-1\)
\(-4+\left|x+5\right|\le y-2\le4-\left|x+5\right|\)
\(-2+\left|x+5\right|\le y\le6-\left|x+5\right|\)
x={-1;-9}=> \(2\le y\le6-4\Rightarrow y=2\)
x={-2;-8}=> \(1\le y\le3\Rightarrow y=1;2;3\)
x={-3;-7} => \(0\le y\le4\Rightarrow y=0;1;2;3;4\)
x={-4;-6}=>\(-1\le y\le5\Rightarrow y=\left\{-1;0;1;2;3;4;5\right\}\)
x={-5}=> \(-2\le y\le6\Rightarrow y=\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
(x;y)=(-1;2); (-2;y={1;2;3}) ....
b) tương tự
=>\(\left(\left|x+5\right|;\left|y-2\right|\right)\in\left\{\left(0;4\right);\left(4;0\right);\left(1;3\right);\left(3;1\right);\left(2;2\right)\right\}\)
TH1: (|x+5|;|y-2|)=(0;4)
=>x+5=0 và |y-2|=4
=>\(\left(x,y\right)\in\left\{\left(-5;6\right);\left(-6;-2\right)\right\}\)
TH2: |x+5|=4 và |y-2|=0
=>\(\left(x,y\right)\in\left\{\left(-1;2\right);\left(-9;2\right)\right\}\)
TH3: |x+5|=1 và |y-2|=3
=>\(\left(x,y\right)\in\left\{\left(-4;5\right);\left(-6;5\right);\left(-4;-1\right);\left(-6;-1\right)\right\}\)
TH4: |x+5|=3 và |y-2|=1
=>(x,y)\(\in\){(-2;3); (-8;3); (-2;1); (-8;1)}
TH5: |x+5|=2; |y-2|=2
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(-3;4\right);\left(-7;4\right);\left(-3;0\right);\left(-7;0\right)\right\}\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
a: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
b: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=5/10+4/10=9/10
a: |x|+|y|<=3
=>(|x|,|y|) thuộc {(0;0); (0;1); (0;2); (0;3); (1;0); (2;0); (3;0); (1;1); (1;2); (2;1)}
=>(x,y) thuộc {(0;0); (0;1); (0;-1); (1;0); (-1;0); (0;2); (2;0); (0;-2); (-2;0); (0;3); (0;-3); (3;0); (-3;0); (1;1); (-1;-1); (-1;1); (1;-1); (1;2); (-1;2); (2;1); (2;-1); (1;-2); (-2;1)}
b: =>(|x+5|,|y-2|) thuộc {(0;4); (4;0); (0;3); (3;0); (0;2); (2;0); (0;1); (1;0); (0;0); (1;1); (1;2); (2;1); (1;3); (3;1); (2;2)}
=>(x+5;y-2) thuộc {(0;4); (0;-4); (4;0); (-4;0); (0;3); (0;-3); (3;0); (-3;0); (0;2); (0;-2); (2;0); (-2;0); (1;0); (-1;0); (0;1); (0;-1); (0;0); (1;1); (1;-1); (-1;1); (-1;-1); (1;2); (2;1); (-1;-2); (-2;-1); (1;-2); (-2;1); (-1;2); (2;-1); (1;3); (-1;-3); (1;-3); (-1;3); (3;1); (-3;-1); (-3;1); (3;-1); (2;2); (-2;-2); (2;-2); (-2;2)}
Đến đây thì dễ rồi, bạn làm như tìm x,y bình thường thôi
`x, y` làm gì có điều kiện đâu ạ?