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HQ
Hà Quang Minh
Giáo viên
12 tháng 1
a)
\(\begin{array}{l}A - C = B\\ \Rightarrow C = A - B \\= 7xy{z^2} - 5x{y^2}z + 3{x^2}yz - xyz + 1 - \left( {7{x^2}yz - 5x{y^2}z + 3xy{z^2} - 2} \right)\\ = 7xy{z^2} - 5x{y^2}z + 3{x^2}yz - xyz + 1 - 7{x^2}yz + 5x{y^2}z - 3xy{z^2} + 2\\ = \left( {7xy{z^2} - 3xy{z^2}} \right) + \left( { - 5x{y^2}z + 5x{y^2}z} \right) + \left( {3{x^2}yz - 7{x^2}yz} \right) - xyz + \left( {1 + 2} \right)\\ = 4xy{z^2} - 4{x^2}yz - xyz + 3\end{array}\)
b)
\(\begin{array}{l}A + D = B\\ \Rightarrow D = B - A \\= - \left( {A - B} \right) = - C \\= - 4xy{z^2} + 4{x^2}yz + xyz - 3.\end{array}\)
c)
\(\begin{array}{l}E - A = B\\ \Rightarrow E = A + B = A \\= 7xy{z^2} - 5x{y^2}z + 3{x^2}yz - xyz + 1 + 7{x^2}yz - 5x{y^2}z + 3xy{z^2} - 2\\ = \left( {7xy{z^2} + 3xy{z^2}} \right) + \left( { - 5x{y^2}z - 5x{y^2}z} \right) + \left( {3{x^2}yz + 7{x^2}yz} \right) - xyz + \left( {1 - 2} \right)\\ = 10xy{z^2} - 10x{y^2}z + 10{x^2}yz - xyz - 1\end{array}\)
B
1
HQ
Hà Quang Minh
Giáo viên
12 tháng 1
\(\begin{array}{l}M - 5{x^2} + xyz = xy + 2{x^2} - 3xyz + 5\\ \Rightarrow M = xy + 2{x^2} - 3xyz + 5 + 5{x^2} - xyz\\ = \left( { - 3xyz - xyz} \right) + \left( {2{x^2} + 5{x^2}} \right) + xy + 5\\ = - 4xyz + 7{x^2} + xy + 5\end{array}\)
28 tháng 10 2021
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
NT
0
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
4 tháng 7 2017
a. \(x^2-2xy+y^2-z^2=\\ \left(x^2-2xy+y^2\right)-z^2\\ =\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
b. \(x^2-6x+9-9y\\ =\left(x^2-6x\right)+\left(9-9y\right)\\ =x\left(x-6\right)+9\left(1-y\right)\)
4 tháng 7 2017
1,
a, \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2\)
=\(\left(x-y-z\right)\left(x-y+z\right)\)
b, hình như sai đề phải ko bn?
Phải là: \(x^2-6x+9-9y^2\)=\(\left(x-3\right)^2-\left(3y\right)^2\)\(=\left(x-3y-3\right)(x+3y-3)\)
c,\(\left(x+y\right)\left(y+z\right)\left(x+z\right)+xyz\)
=\((x^2+xy+yz+xz)\left(z+y\right)+xyz\)
=\((x^2z+xyz+xz^2)+(x^2y+xy^2+xyz)+\)\(\left(yz^2+y^2z+xyz\right)\)
= xz(x + y + z) + xy(x + y + z) + yz(x + y +z)
=(x+y+z)(xz+xy+yz)
2,a,\(\left(x-2\right)\left(x+1\right)=0\)\(\Rightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b,Đề phải thế này nha:
\(5x\left(x-3\right)-x+3=0\)\(\Rightarrow\)(x - 3)(5x - 1)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
Chúc các bn học tốt 
12 tháng 8 2018
a) \(x^2-3xy+x-3y=x\left(x-3y\right)+\left(x-3y\right)=\left(x-3y\right)\left(x+1\right)\)
b) \(x^2-6x-y^2+9=x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
c) \(7x^3y-14x^2y+7xy=7xy\left(x^2-2x+1\right)=7xy\left(x-1\right)^2\)
12 tháng 8 2018
\(x^2-3xy+x-3y=\left(x^2+x\right)-\left(3xy+3y\right)=x\left(x+1\right)-3y\left(x+1\right)=\left(x+1\right)\left(x-3y\right)\)
\(x^2-6x-y^2+9=\left(x^2-2.x.3+3^2\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
\(7x^3y-14x^2y+7xy=\left(7x^3y-7x^2y\right)-\left(7x^2y-7xy\right)=7x^2y.\left(x-1\right)-7xy.\left(x-1\right)\)
\(=\left(x-1\right).\left(7x^2y-7xy\right)=7xy.\left(x-1\right).\left(x-1\right)=7xy.\left(x-1\right)^2\)
`Answer:`
`m-3xyz+5x^2-7xy+9=6x^2+xyz+2xy+3-y^2`
`<=>m=(6x^2+xyz+2xy+3-y^2)+(3xyz-5x^2+7xy-9)`
`<=>(xyz+3xyz)+(6x^2-5x^2)+(2xy+7xy)-y^2+(3-9)`
`<=>m=4xyz+x^2+9xy-y^2-6`