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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
Bài 1 :
a/ \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy....
b/ \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-9x-x+9=0\)
\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy...
c/ \(x^2+9x+8=0\)
\(\Leftrightarrow x^2+8x+x+8=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)
Vậy ...
d/ \(x^2-11x+10=0\)
\(\Leftrightarrow x^2-11x+10=0\)
\(\Leftrightarrow x^2-x-10x+10=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
Vậy...
Bài 2 :
Ta có :
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy....
Bài 3 : không hiểu đề lắm ???!!!!
Bài 4 :
Ta có :
\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)
Thay (1) ta có :
\(\frac{x}{y}=16\)
\(\Leftrightarrow\frac{2y^2}{y}=16\)
\(\Leftrightarrow2y=16\)
\(\Leftrightarrow y=8\Leftrightarrow x=128\)
Vậy...
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
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: \(=\left(x+y\right)^2-4\left(x+y\right)+1=9-12+1=-2\)
