Cho \(x,y\in N,y>x\)t/m \(\left(2y-1\right)^2=\left(2y-x\right)\left(6y+x\right)\).C/M : 2y-x là số chính phương
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T
8 tháng 12 2019
1/ b) Đặt \(\sqrt[3]{6x+4}=a\Rightarrow a^3=6x+4\)
Ta có hệ: \(\left\{{}\begin{matrix}x^3=6a+4\\a^3=6x+4\end{matrix}\right.\)
Lấy pt trên trừ pt dưới vế với vế, suy ra:
\(\left(x-a\right)\left(x^2+ax+a^2+6\right)=0\)
\(\Leftrightarrow x=a\Leftrightarrow x^3-6x-4=0\Leftrightarrow\left(x+2\right)\left(x^2-2x-2\right)=0\)
22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
E
26 tháng 10 2018
Xin câu a :3
a) (x + y + 1)2 = 3(x2 + y2) + 1
<=> x2 + y2 + 1 + 2xy + 2x + 2y = 3x2 + 3y2 + 1
<=> 2x2 + 2y2 - 2xy - 2x - 2y = 0
<=> (x2 - 2xy + y2) + (x2 - 2x + 1) + (y2 - 2y + 1) = 2
<=> (x - y)2 + (x - 1)2 + (y - 1)2 = 2
Vì 2 = 02 + 12 + 12 nên ta có các TH sau:
TH1:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=1\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=0\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=0\\x=1;y=2\end{matrix}\right.\)
TH3:
\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=1\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=0;y=1\end{matrix}\right.\)
Vậy ...
26 tháng 10 2018
a) ta có : \(\left(x+y+1\right)^2=3\left(x^2+y^2\right)+1\)
\(\Leftrightarrow x^2+y^2+1+2xy+2y+2x=3x^2+3y^2+1\)
\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2=\left(x-y\right)^2-2\le0\)
\(\Leftrightarrow-\sqrt{2}\le x-y\le\sqrt{2}\) --> ...
b) \(\left(2x-y-2\right)^2=7\left(x-2y-y^2-1\right)\)
\(\Leftrightarrow4x^2+y^2+4-4xy+4y-4x=7x-14y-7y^2-7\)
\(\Leftrightarrow2x^2-4xy+2y^2+2x^2-11x+\dfrac{121}{16}+6y^2+18y+\dfrac{9}{4}=\dfrac{-19}{16}\left(vl\right)\)
câu c tương tự .
4 tháng 12 2019
Ta có: \(\left( 1 \right) \Leftrightarrow \left( {1 - y} \right)\sqrt {x - y} + \left( {x - y - 1} \right) + y - 1 = \left( {x - y - 1} \right)\sqrt y \)
\( \Leftrightarrow \left( {1 - y} \right)\left( {\sqrt {x - y} - 1} \right) + \left( {x - y - 1} \right)\left( {1 - \sqrt y } \right) = 0\\ \Leftrightarrow \left( {1 - \sqrt y } \right)\left( {1 + \sqrt y } \right)\left( {\sqrt {x - y} - 1} \right) + \left( {\sqrt {x - y} - 1} \right)\left( {\sqrt {x - y} + 1} \right)\left( {1 - \sqrt y } \right) = 0\\ \Leftrightarrow \left( {1 - \sqrt y } \right)\left( {\sqrt {x - y} - 1} \right)\left( {1 + \sqrt y + \sqrt {x - y} + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt y = 1\\ \sqrt {x - y} = 1\\ 2 + \sqrt y + \sqrt {x - y} = 0 \text{(vô nghiệm do vế trái dương)} \end{array} \right. \)
\(\Leftrightarrow y = 1 \vee x = y + 1 \)
* Với \(y=1\) thay vào (2) ta được \(-3x+9=0 \Leftrightarrow x = 3\)
Vậy nghiệm hệ phương trình là \((3;1)\)
* Với \(x=y+1\) thay vào (2) ta được:
\( 2{y^2} - 3\left( {y + 1} \right) + 6y + 1 = 2\sqrt {y + 1 - 2y} - \sqrt {4\left( {1 + y} \right) - 5y - 3} \\ \Leftrightarrow 2{y^2} + 3y - 2 = \sqrt {1 - y} \left( * \right) (ĐK: y \in \left[ {0;1} \right]) \)
\( \Leftrightarrow 2\left( {{y^2} + y - 1} \right) = \sqrt {1 - y} - y \Leftrightarrow 2\left( {{y^2} + y - 1} \right) = \dfrac{{1 - y - {y^2}}}{{\sqrt {11 - y} + y}}\\ \Leftrightarrow \left( {{y^2} + y - 1} \right)\left( {2 + \dfrac{1}{{\sqrt {1 - y} + y}}} \right) = 0\\ \Leftrightarrow {y^2} + y - 1 = 0\left( {do2 + \dfrac{1}{{\sqrt {1 - y} + y}} > 0\forall y \in \left[ {0;1} \right]} \right)\\ \Leftrightarrow y = \dfrac{{ - 1 + \sqrt 5 }}{2} \)
Vậy nghiệm hệ phương trình là: \(\left( {\dfrac{{1 + \sqrt 5 }}{2};\dfrac{{ - 1 + \sqrt 5 }}{2}} \right) \)
28 tháng 10 2020
a. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)