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21 tháng 7 2023

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HQ
Hà Quang Minh
Giáo viên
12 tháng 1

a)      \({\left( {x - 3y} \right)^2} - {\left( {x + 3y} \right)^2} = \left( {x - 3y + x + 3y} \right).\left( {x - 3y - x - 3y} \right) = \left( {2x} \right).\left( { - 6y} \right) =  - 12xy\)

b)       

\(\begin{array}{l}{\left( {3x + 4y} \right)^2} + {\left( {4x - 3y} \right)^2} = {\left( {3x} \right)^2} + 2.3x.4y + {\left( {4y} \right)^2} + {\left( {4x} \right)^2} - 2.4x.3y + {\left( {3y} \right)^2}\\ = 9{x^2} + 24xy + 16{y^2} + 16{x^2} - 24xy + 9{y^2}\\ = \left( {9{x^2} + 16{x^2}} \right) + \left( {24xy - 24xy} \right) + \left( {16{y^2} + 9{y^2}} \right)\\ = 25{x^2} + 25{y^2}\end{array}\)

b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)

\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)

\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)

\(=-x^2+18xy\)

c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)

\(=\left(2a-3b\right)^2-16c^2\)

\(=4a^2-12ab+9b^2-16c^2\)

12 tháng 7 2017

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

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

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

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

\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=4y^3+y^2+6xy^2\)

12 tháng 7 2017

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-\left(9x^2y-12y^2\right)\)

\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y+y^2+y^3+y\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

Chúc bạn học tốt!!!

12 tháng 7 2017

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

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

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y\left(1-y-y^2-1\right)\)

\(=9y^3-y\left(-y-y^2\right)\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

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: \(x^2-y^2-2x+2y\)

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

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

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

27 tháng 9 2019

nhân ra hết là đc

27 tháng 9 2019

kệ chẳng quan tâm

16 tháng 6 2017

Bài 1 :

a) \(\left(3x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=2014\)

\(\Leftrightarrow9x^2-6x+1-\left(9x^2-4\right)=2014\)

\(\Leftrightarrow-6x=2009\)

\(\Leftrightarrow x=-\dfrac{2009}{6}=-334\dfrac{5}{6}\)

b) \(5x^2+4xy+4y^2+4x+1=0\)

\(\Leftrightarrow\left(x^2+4xy+4y^2\right)+\left(4x^2+4x+1\right)=0\)

\(\Leftrightarrow\left(x+2y\right)^2+\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2y=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{4}\end{matrix}\right.\)

Bài 2 :

Ta có :

\(D=\left(4x^2-12xy+9y^2\right)-\left(9y^2-4\right)-\left(1-4x+4x^2\right)+12xy-4x\)

\(=4x^2-12xy+9y^2-9y^2+4-1+4x-4x^2+12xy-4x=3\)

Vậy biểu thức D không phụ thuộc vào các biến x,y

17 tháng 8 2020

a) \(4x^2\left(5x^3-2x+3\right)\)

\(=20x^5-8x^3+12x^2\)

b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)

\(=12y^5+2y^4-y^2\)

c) \(\left(5x^2-4x\right)\left(x-2\right)\)

\(=5x^3-14x^2+8x\)

d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x^2+22x-55-6x^2-23x-21\)

\(=-x-76\)

17 tháng 8 2020

1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)

2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)

3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)

4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)

\(=6x^2+23x-55-6x^2-23x-21=-76\)

12 tháng 1 2019

a) Đề sai nha bạn :) mấy dấu cộng bạn phỉa chuyển thành dấu nhân nhé

\(A=\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)

\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)

\(A=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)+1\)

\(A=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)

\(A=2^{512}-1+1\)

\(A=2^{512}\)

12 tháng 1 2019

b . ( 5x - 3y + 4z )( 5x - 3y - 4z ) = ( 5x - 3y )^2 - ( 4z )^2 = 25x^2 - 30xy + 9y^2 - 16z^2 = 25( y^2 + z^2 ) - 30xy + 9y^2 - 16z^2 = 9z^2 + 34y^2 - 30xy ( 1 )

      ( 3x - 5y )^2 = 9x^2 - 30xy + 25y^2 = 9( y^2 + z^2 ) - 30xy + 25y^2 = 34y^2 + 9z^2 - 30xy ( 2 )

Tu ( 1 ) va ( 2 ) => dpcm