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bài 1:

a) x2 + 10x + 26 + y2 + 2y

= (x2 + 10x + 25) + (y2 + 2y + 1)

= (x + 5)2 + (y + 1)2

b) z2 - 6z + 5 - t2 - 4t

= (z - 3)2 - (t + 2)2

c) x2 - 2xy + 2y2 + 2y + 1

= (x2 - 2xy + y2) + (y2 + 2y + 1)

= (x - y)2 + (y + 1)2

d) 4x2 - 12x - y2 + 2y + 1

= (4x2 - 12x ) - (y2 + 2y + 1)

= ......................................

ok mk nhé!! 4545454654654765765767587876968345232513546546575675767867876876877687975675

19 tháng 8 2020

a) \(x^2+10x+26+y^2+2y\)

\(=x^2+2.5x+25+1+y^2+2y\)

\(=\left(x^2+2.5x+25\right)+\left(1+2y+y^2\right)\)

\(=\left(x+5\right)^2+\left(1+y\right)^2\)

b) \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)\)

\(=\left(x-y\right)^2+\left(y+1\right)^2\)

c) \(z^2-6z+13+t^2+4t\)

\(=z^2-2.3z+9+4+t^2+4t\)

\(=\left(z^2-2.3x+9\right)+\left(4+4t+t^2\right)\)

\(=\left(z-3\right)^2+\left(2+t\right)^2\)

d) \(4x^2+2z^2-4xz-2z+1\)

\(=4x^2+z^2+z^2-4xz-2z+1\)

\(=\left(4x^2-4xz+z^2\right)+\left(z^2-2z+1\right)\)

\(=\left(2x-z\right)^2+\left(z-1\right)^2\)

10 tháng 6 2018

a) \(\Rightarrow\left(x^2+2\times5x+25\right)+\left(y^2+2y+1\right)\)

\(\Rightarrow\left(x+5\right)^2+\left(y+1\right)^2\)

9 tháng 8 2021

câu d thì s ạ ?

25 tháng 7 2015

c)  x2-2xy+2y2+2y+1

=(x2-2xy+y2)+(y2+2y+1)

=(x-y)2+(y+1)2

d)   4x2-12x-y2+2y+8

=(4x2-12x+9)-(y2+2y+1)

=(2x-3)2-(y-1)2

 

25 tháng 7 2015

a)  x2+10x+26+y2+2y

=(x2+10x+25)+(1+y2+2y)

=(x+5)2+(y+1)2

b)z2-6z+9-4-t2-4t

=(z-3)2-(t+2)2

 

1 tháng 10 2018

A=\(x^3-2x^2+x\)

=x.(x2-2x+1)

=x(x-1)2

B=\(2x^2+4x+2-2y^2\)

=\(2\left(x^2+2x+1-y^2\right)\)

=\(2.\left[\left(x+1\right)^1-y^2\right]\)

=\(2\left(x+1-y\right)\left(x+1+y\right)\)

C=\(2xy-x^2-y^2+16\)

=\(-\left(-2xy+x^2+y^2-16\right)\)

=\(-\left[\left(x-y\right)^2-4^2\right]\)

=-(x-y-4)(x-y+4)

D=\(x^3+2x^2y+xy^2-9x\)

=\(x\left(x^2+2xy-y^2-9\right)\)

=\(x.\left[\left(x-y\right)^2-3^2\right]\)

=x.(x-y-3)(x-y+3)

E=\(2x-2y-x^2+2xy-y^2\)

\(=\left(2x-2y\right)-\left(x^2-2xy+y^2\right)\)

=\(2\left(x-y\right)-\left(x-y\right)\left(x-y\right)\)

=(x-y)(2x-2y-x+y)

=(x-y)(x+y)

1 tháng 10 2018

ở câu B:

(x+1)^1 sửa giùm mk thành (x+1)^2

a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)

b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)

13 tháng 10 2018

a/ \(x^2-5x+5y-y^2=\left(x^2-y^2\right)-\left(5x-5y\right)=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)

b/ \(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x^2-2xy+y^2\right)-\left(2x\right)^2\right]=3\left[\left(x-y\right)^2-\left(2x\right)^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

c/ \(x^2-2xy+y^2-xz+yz=\left(x^2-2xy+y^2\right)-\left(xz-yz\right)=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

d/ \(x^2-x+2y-4y^2=\left(x^2-4y^2\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y\right)-\left(x+2y\right)=\left(x+2y\right)\left(x-2y-1\right)\)

e/ \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

13 tháng 10 2018

a) x2 - 5x + 5y - y2

= ( x2 - y2 ) - ( 5x - 5y )

= ( x - y )( x + y ) - 5( x - y )

= ( x - y )( x + y - 5 )

b) 3x2 - 6xy + 3y2 - 12z2

= 3( x2 - 2xy + y2 - 4z2 )

= 3[( x2 - 2xy + y2 ) - 4z2 ]

= 3[( x - y )2 - 4z2 ]

= 3( x - y - 2z )( x - y + 2z )

c) x2 - 2xy + y2 - xz - yz

= ( x2 - 2xy + y2 ) - ( xz - yz )

= ( x - y )2 - z( x - y )

= ( x - y )( x - y - z )

d) x2 - x + 2y - 4y2

= ( x2 - 4y2 ) - ( x - 2y )

= ( x - 2y )( x + 2y ) - ( x - 2y )

= ( x - 2y )(x + 2y - 1 )

e) x6 - y6

= ( x3 )2 - ( y3 )2

= ( x3 - y3 )( x3 + y3 )

= ( x - y )( x2 + xy + y2 )( x + y )( x2 - xy + y2 )

Chúc bạn học tốt hihi

5 tháng 11 2017

\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)

\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)

5 tháng 11 2017

Hỏi đáp ToánHỏi đáp ToánHỏi đáp Toán

4 tháng 12 2018

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

\(=-2x^2+2x+6\)

\(=-2\left(x^2-x-3\right)\)

b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)

\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)

\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)

\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)

\(=x^4+4x^2+4-x^4+16\)

\(=4x^2+20\)

\(=4\left(x^2+5\right)\)

c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)

\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)

\(=x^6-3x^4+3x^2-1-x^6+1\)

\(=-3x^4+3x^2\)

\(=-3x^2\left(x^2-1\right)\)

\(=-3x^2\left(x-1\right)\left(x+1\right)\)

e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)

\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)

\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)

\(=\left(2x-1-2x-1\right)^2\)

\(=\left(-2\right)^2=4\)

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)

\(=\left(x-y+z+y+z\right)^2\)

\(=\left(x+2z\right)^2\)

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)

\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)

\(=\left(2x+3-2x-5\right)^2\)

\(=\left(-2\right)^2=4\)

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)

\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)

\(=5x^2+2x^2+3x-1-3x-3\)

\(=7x^2-4\)