Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.a)\(20x-5y=5\left(4x-y\right)\)
b)\(5x\left(x-1\right)-3x\left(x-1\right)=\left(5x-3x\right)\left(x-1\right)=2x\left(x-1\right)\)
c)\(x\left(x+y\right)-6x-6y=x\left(x+y\right)-6\left(x+y\right)=\left(x-6\right)\left(x+y\right)\)
d)\(6x^3-9x^2=3x^2\left(2x-3\right)\)
e)\(4x^2y-8xy^2+10x^2y^2=2xy\left(2x-8y+10xy\right)\)
g)\(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
h)\(8x^4+12x^2y-16x^3y^4=4x^2\left(2x^2+12y-16xy^4\right)\)
2.a)\(3x\left(x+1\right)-5y\left(x+1\right)=\left(3x-5y\right)\left(x+1\right)\)
b)\(3x\left(x-6\right)-2\left(x-6\right)=\left(3x-2\right)\left(x-6\right)\)
c)\(4y\left(x-1\right)-\left(1-x\right)=4y\left(x-1\right)+\left(x-1\right)=\left(4y+1\right)\left(x-1\right)\)
d)\(\left(x-3\right)^3+3-x=\left(x-3\right)^3-\left(x-3\right)=\left(x-3\right)\left[\left(x-3\right)^2-1\right]=\left(x-3\right)\left(x-2\right)\left(x-4\right)\)
e)\(7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(7x+1\right)\left(x-y\right)\)
h)\(3x^3\left(2y-3z\right)-15x\left(2y-3z\right)^2=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]\)
k)Sai đề: \(3x\left(z+2\right)+5\left(-z-2\right)=3x\left(z+2\right)-5\left(z+2\right)=\left(3x-5\right)\left(z+2\right)\)
l)\(18x^2\left(3+x\right)+3\left(x+3\right)=3\left(x+3\right)\left(6x^2+1\right)\)
m)\(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)
n)\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
Bài 1 :
a) (3a+4b)3+(3a-4b)3-48a2b2
=27a3+108a2b+144ab2+64b3+27a3-108a2b+144ab2-64b3-48a2b2
=54a3+288ab2-48a2b2
=2a(27a2+144b2-24ab)
b) (5x+2y)(5x-2y)+(2x-y)3+(2x+y)3
=25x2-4y2+8x3-12x2y+6xy2-y3+8x3+12x2y+6xy2+y3
=16x3+25x2-y2+12xy2
=x2(16x+25)-y2(1-12x)
Bài 2 :
\(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
b)\(x^3-4x^2+3x=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{3}\\x=1\end{cases}}\)
c)Nếu đề đổi thành =1 thì có vẻ hợp lí hơn
d)\(\left(3x-1\right)^3-3\left(3x+2\right)^2+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-3\left(9x^2+12x+4\right)+13=0\)
\(\Leftrightarrow27x^3-27x^2+9x-1-27x^2-36x-12+13=0\)
\(\Leftrightarrow27x^3-54x^2-27x=0\)
\(\Leftrightarrow27x\left(x^2-2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}27x=0\\x^2-2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\-\left(x^2+2x+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\-\left(x+1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
#H
a) 5x-15y=5x-3.5.y=5(x-3y)
c) 14xy(xy+28x)
d) \(\dfrac{2}{7}\left(3x-1\right)\left(x-1\right)\)
e) (x-1)3
f) (x+y-2x)(x+y+2x)=(y-x)(3x+y)
g) (3x+\(\dfrac{1}{2}\))(9x2+\(\dfrac{3}{2}x\)+\(\dfrac{1}{4}\))
h) (x+y-x+y)\(\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
2a)
(x+1)(x2+2x)=0
(x+1)x(x+2)=0
\(\left[{}\begin{matrix}x+1=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)