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a/ |2x - 3| + |y - 2| = 0
Vì: \(\left\{{}\begin{matrix}\left|2x-3\right|\ge0\forall x\\\left|y-2\right|\ge0\forall y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x-3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=2\end{matrix}\right.\)
b/ |3x - 4| + |x - y| = 0
Vì: \(\left\{{}\begin{matrix}\left|3x-4\right|\ge0\forall x\\\left|x-y\right|\ge0\forall x;y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3x-4=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\x=y=\dfrac{4}{3}\end{matrix}\right.\)
Vậy x = y = 4/3
c/ \(\left|2x+y-1\right|+\left|2y-3\right|=0\)
Vì: \(\left\{{}\begin{matrix}\left|2x+y-1\right|\ge0\forall x;y\\\left|2y-3\right|\ge0\forall y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x+y-1=0\\2y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-1=-y\\y=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=-\dfrac{3}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..........
d/ \(\left|x+y-5\right|+\left|2x-y+8\right|=0\)
Vì: \(\left\{{}\begin{matrix}\left|x+y-5\right|\ge0\\\left|2x-y+8\right|\ge0\end{matrix}\right.\)∀x;y
=> \(\left\{{}\begin{matrix}x+y-5=0\\2x-y+8=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+y=5\\2x-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\2\left(5-y\right)-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\10-2y-y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\-3y=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5-6=-1\\y=6\end{matrix}\right.\)
Vậy x = -1; y = 6
a/ |2x - 3| + |y - 2| = 0
Vì: {|2x−3|≥0∀x|y−2|≥0∀y{|2x−3|≥0∀x|y−2|≥0∀y
=> {2x−3=0y−2=0⇒⎧⎨⎩x=32y=2{2x−3=0y−2=0⇒{x=32y=2
b/ |3x - 4| + |x - y| = 0
Vì: {|3x−4|≥0∀x|x−y|≥0∀x;y{|3x−4|≥0∀x|x−y|≥0∀x;y
=> {3x−4=0x−y=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=43x=y=43{3x−4=0x−y=0⇔{x=43x=y=43
Vậy x = y = 4/3
c/ |2x+y−1|+|2y−3|=0|2x+y−1|+|2y−3|=0
Vì: {|2x+y−1|≥0∀x;y|2y−3|≥0∀y{|2x+y−1|≥0∀x;y|2y−3|≥0∀y
=> {2x+y−1=02y−3=0⇔⎧⎨⎩2x−1=−yy=32{2x+y−1=02y−3=0⇔{2x−1=−yy=32
⇔⎧⎪ ⎪⎨⎪ ⎪⎩2x−1=−32y=32⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=−14y=32⇔{2x−1=−32y=32⇔{x=−14y=32
Vậy..........
d/ |x+y−5|+|2x−y+8|=0|x+y−5|+|2x−y+8|=0
Vì: {|x+y−5|≥0|2x−y+8|≥0{|x+y−5|≥0|2x−y+8|≥0∀x;y
=> {x+y−5=02x−y+8=0{x+y−5=02x−y+8=0⇔{x+y=52x−y=−8⇔{x+y=52x−y=−8
⇔{x=5−y2(5−y)−y=−8⇔{x=5−y2(5−y)−y=−8
⇔{x=5−y10−2y−y=−8⇔{x=5−y10−2y−y=−8
⇔{x=5−y−3y=−18⇔{x=5−yy=6⇔{x=5−6=−1y=6⇔{x=5−y−3y=−18⇔{x=5−yy=6⇔{x=5−6=−1y=6
Vậy x = -1; y = 6
CHÚC BẠN HỌC TỐT
a)Ta có : 2x+2y-z-7=0 => 2x+2y-z=7
Ta có : \(x=\frac{y}{2}=>\frac{x}{2}=\frac{y}{4}\)
Mà \(\frac{y}{4}=\frac{z}{5}\)nên \(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}=\frac{2x+2y-z}{4+8-5}=\frac{7}{7}=1\)
Từ \(\frac{x}{2}=1=>x=2\)
Từ\(\frac{y}{4}=1=>y=4\)
Từ \(\frac{z}{5}=1=>z=5\)
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)