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Bài 2:
a) \(=x^2-36y^2\)
b) \(=x^3-8\)
Bài 3:
a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)
b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính :
a) ( x - 2y )( 3xy + 6y^2 + x )
b) [ 4 ( x-y )^5 + 2( x - y )^3 - 3( x - y )^2 ] : ( y - x )^2
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(x-2y\right)\left(3xy+6y^2+x\right)=x\left(3xy+6y^2+x\right)-2y\left(3xy+6y^2+x\right)\)\(=3x^2y+6xy^2+x^2-6xy^2-12y^3-2xy\)
\(=3x^2y+x^2-12y^3-2xy\)
b)\(\text{[}4\left(x-y\right)^5+2\left(x-y\right)^3-3\left(x-y\right)^2\text{]}:\left(y-x\right)^2\)
=\(\text{[}4\left(x-y\right)^5+2\left(x-y\right)^3-3\left(x-y\right)^2\text{]}:\left(x-y\right)^2\)
=\(^{=4\left(x-y\right)^3+2\left(x-y\right)-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có a + b + 8 = 0
=> x3 + 3x2 + 6x + y3 + 3y2 + 6y + 8 = 0
=> (x3 + 3x2 + 3x + 1) + (y3 + 3y2 + 3y + 1) + (3x + 3y + 6) = 0
=> (x + 1)3 + (y + 1)3 + 3(x + y + 2) = 0
=> (x + y + 2)[(x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3] = 0
Vì (x + 1)2 + (x + 1)(y + 1) + (y + 1)2 + 3 \(>0\forall x;y\)
=> x + y + 2 = 0
=> x + y = -2
Vậy A = -2
xyz bạn ơi! tại sao từ dòng 3 lại thành dòng 4 vậy
thank you bạn!!! <3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)
\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)
\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
Bài 2:
\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)
bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)
nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)
Thay \(x=4;y=1\) vào \(P\), ta được:
\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)
\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)
\(=1-1=0\)
Vậy \(P=0\) khi \(x=4;y=1\).
![](https://rs.olm.vn/images/avt/0.png?1311)
(x2 -y2 +6y -9) : (x-y+3)
= [x2-(y2-6y+9)] : (x-y+3)
=[x2-(y-3)2 ] : (x-y+3)
=[(x-y+3)(x+y-3)] :(x-y+3)
=x+y-3
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow\left(x+\frac{1}{6}y+3\right)^2=x^2+\left(\frac{y}{6}\right)^2+3^2+2.x.\frac{y}{6}+2.\frac{y}{6}.3+2.x.3\)
\(=x^2+\frac{y^2}{36}+9+\frac{xy}{3}+y+6x\)
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