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9 tháng 7 2019

1.

a)

\(5x\left(x-2y\right)+2\left(2y-x\right)^2\\ =5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left[5x+2\left(x-2y\right)\right]\\ =\left(x-2y\right)\left(5x+2x-4y\right)\\ =\left(x-2y\right)\left(7x-4y\right)\)

b)

\(7x\left(y-4\right)^2-\left(4-y\right)^3\\ =7x\left(4-y\right)^2-\left(4-y\right)^3\\ =\left(4-y\right)^2\left(7x+y-4\right)\)

c)

\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\\ =\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\\ =\left(4x-8\right)\left(x^2+6-x-7-9\right)\\ =4\left(x-2\right)\left(x^2-x-10\right)\)

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: \(x^2-y^2-2x+2y\)

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

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

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

AH
Akai Haruma
Giáo viên
14 tháng 8 2019

a)

\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)

\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)

b)

\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)

c)

\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)

\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)

\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)

d)

\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)

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

e)

\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)

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

AH
Akai Haruma
Giáo viên
12 tháng 8 2019

a)

\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)

\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)

b)

\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)

c)

\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)

\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)

\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)

d)

\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)

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

e)

\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)

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

26 tháng 11 2021

Answer:

Câu đầu bạn xem lại.

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

\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)

\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)

\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)

\(=29\)

\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)

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

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

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

\(=2y^3\)

18 tháng 2 2020

Bài 2 :

a) \(\left(5x^2y-8xy^2+y^3\right)\left(2x^3+x^2y-3y^2\right)\)

\(=10x^5y+5x^4y^2-15x^2y^3-16x^4y^2-8x^3y^3+24xy^4+2x^3y^3+x^2y^4-3y^5\)

\(=10x^5y-11x^4y^2-6x^3y^3+x^2y^4-15x^2y^3+24xy^4-3y^5\)

26 tháng 12 2021

tách nhỏ câu hỏi ra bạn

26 tháng 12 2021

\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)

\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)

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

\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

 

2 tháng 1 2022

đáp án: a là đúng

20 tháng 11 2021

A