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19 tháng 6 2022
1: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+2y^2\)
\(=\left(x^2+y^2\right)\left(3x^2+2y^2\right)+2y^2\)
\(=3x^2+2y^2+2y^2=3x^2+4y^2\)
2: \(=7\left(x-y\right)+4a\left(x-y\right)-5\)
=-5
3: \(=\left(x-y\right)\left(x^2+xy+y^2\right)-xy\left(x-y\right)+3=3\)
4: \(=\left(x+y\right)^2-4\left(x+y\right)+1=9-12+1=-2\)
7 tháng 8 2017
bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
AT
7 tháng 7 2018
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
LV
28 tháng 9 2018
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
5 tháng 7 2017
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-\frac{1}{2}=0\\\frac{1}{2}y+\frac{3}{5}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=\frac{-6}{5}\end{cases}}\)
b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{5}{y}y-\frac{1}{2}\right|< 0\)
Vì trị tuyệt đối của một giá trị luôn lớn hơn hoặc bằng 0
Mà đề cho tổng trên nhỏ hơn 0
=> Không thể làm :v
Ta có: \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left(y-2\right)^{2028}\ge0\end{matrix}\right.\)
=> \(\left|x+1\right|+\left(y-2\right)^{2028}\ge0\)
Dấu = xảy ra khi:
\(\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Khi đó:
`A =` \(x^{2024}+\left(5-y\right)^3\)
`=` \(\left(-1\right)^{2024}\) `+ (5-2)^3 `
`= 1 + 3^3 `
`=1 + 27`
`= 28`
Vậy `A = 28`
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