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8 tháng 8 2018

a) \(2x+2y\)

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

b) \(5x+20y\)

\(=5\left(x+4y\right)\)

c) \(6xy-30y\)

\(=6y\left(x-5\right)\)

d) \(5x\left[x-110-10y\left(x-11\right)\right]\)

\(=5x\left(x-110-10xy+110\right)\)

\(=5x\left(x-10xy\right)\)

\(=5x^2\left(1-10y\right)\)

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

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

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

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

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

h) \(5x\left(x-2y\right)+2\left(2y-x\right)\)

\(=5x\left(x-2y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x-2\right)\)

i) \(x^2y^3-\dfrac{1}{2}x^4y^8\)

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

j) \(a^2b^4+a^3b-abc\)

\(=ab\left(ab^3+a^2-c\right)\)

18 tháng 1 2018

Chị cũng là fan của BTS à

18 tháng 1 2018

Chị hâm mộ V đúng không

3 tháng 10 2017

Bài 1 câu g bạn kia làm sai mình sửa lại nhá

\(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a^2-2ab+b^2\right)-12c^2\)

\(=3\left(a-b\right)^2-12c^2\)

\(=3\left[\left(a-b\right)^2-4c^2\right]\)

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

3 tháng 10 2017

Để mình làm tiếp cho :))

Bài 2 :

Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)

\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)

\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)

\(=37,5.10-7,5.10\)

\(=10.30=300\)

Câu b : \(35^2+40^2-25^2+80.35\)

\(=\left(35^2+80.35+40^2\right)-25^2\)

\(=\left(30+45\right)^2-25^2\)

\(=75^2-25^2\)

\(=\left(75+25\right)\left(75-25\right)\)

\(=100.50=5000\)

Bài 3 :

Câu a : \(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)

Câu b : \(2x-2y-x^2+2xy-y^2=0\)

\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)

Câu c :

\(x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(x^2\left(x-3\right)+27-9x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)

Bài 4 :

Câu a :

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

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

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

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

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

Câu b :

\(x^2+x-6\)

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

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

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

Câu c :

\(x^2-5x+6\)

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

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

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

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

Câu d :

\(x^4+4\)

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

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

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

4 tháng 8 2019

a. x4 + x2y2 + y4 = (x4 + 2x2y2 + y4) - x2y2

= (x2 + y2)2 – (xy)2

= [(x2 + y2) + xy] [(x2 + y2) – xy]

= (x2 + xy + y2)(x2 –xy + y2)

Bài 1:

a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)

b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)

c)Đề sai hoàn toàn

d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)

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

g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)

i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)

d)5.(x-y)-y(x-y)

=(x-y)(5-y)

e) y.(x-z)+7(z-x)

=y.(x-z)-7(x-z)

=(x-z)(y-7)