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Bài 2:
a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)
\(=7^3+2\cdot7^2=441\)
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B1 : a, M = x3-3xy(x-y)-y3-x2+2xy-y2
= ( x3-y3)-3xy(x-y) -(x2-2xy+y2)
= (x-y)(x2+xy+y2)-3xy(x-y)-(x-y)2
= (x-y) [(x2+xy+y2-3xy-(x-y)]
= (x-y)[(x2-2xy+y2)-(x-y)
= (x-y)[(x-y)2-(x-y)]
= (x-y)(x-y)(x-y-1)
= (x-y)2(x-y-1)
= 72(7-1) = 49 . 6= 294
N = x2(x+1)-y2(y-1)+xy-3xy(x-y+1)-95
= x3+x2-(y3-y2)+xy-(3x2y-3xy2+3xy)-95
= x3+x2-y3+y2+xy-3x2y+3xy2-3xy-95
= (x3-y3)+(x2-2xy+y2)-(3x2y+y2)-(3x2y-3xy2)-95
=(x-y)(x2+xy+y2)+(x-y)2-3xy(x-y)-95
= (x-y)(x2+xy+y2+x-y-3xy)-95
= (x-y)[(x2-2xy+y2)+(x-y)]-95
= (x-y)[(x-y)2+(x-y)]-95
=(x-y)(x-y)(x-y+1)-95
= (x-y)2(x-y+1)-95
= 72(7+1)-95=297
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\(a)\)
\(21\left(x+3\right)^3:\left(3x+9\right)^2\)
\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)
\(=7\left(x+3\right):3\)
Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)
\(b)\)
Thay vào ta được:
\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)
\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)
\(=1^4:\left(1^3.1\right)\)
\(=1:1\)
\(=1\)
\(c)\)
Thay vào ta được:
\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)
\(=-6.10.7\)
\(=-420\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)
\(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)
\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)
\(=\frac{x+y-z}{x-y+z}\)
Ta thay : \(x=0;y=2009;z=2010\) ta được :
\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)
Chúc bạn học tốt !!!
\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)
\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)
Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :
\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)
`M=2(x^3 -y^3 )-3(x^2 +y^2)`
`M=2(x-y)(x^2 +xy+y^2 )-3x^2 -3y^2`
`M=2x^2 +2xy+2y^2 -3x^2 -3y^2`
`M=-x^2 +2xy-y^2`
`M=-(x^2 -2xy+y^2)`
`M=-(x-y)^2`
`M=-(1)^2`
`M=-1`
\(M=2\left(x^3-y^3\right)-3\left(x^2-y^2\right)\)
\(M=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\left(x+y\right)\)
\(M=2\left[x^2+x\left(x-1\right)+\left(x-1\right)^2\right]-3\left(2x-1\right)\)
\(M=2\left(x^2+x^2-x+x^2-2x+1\right)-6x+3\)
\(M=6x^2-12x+5\)
Đề bài yêu cầu tính giá trị nhưng mình cũng không rõ là giá trị gì nên mình làm đến đây thôi nhé.