Bài 1 Phân tích mỗi đa thức sau thanh nhân tử
a)5x-15y
b)12y(2x-5y)+6xy(5-2x)
c)x^2-7x+12
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10 tháng 11 2023
a: \(P=x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b: \(P=3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)\)
\(=\left(x+5\right)\left(3x-1\right)\)
c: \(P=-2x^2-7x-5\)
\(=-\left(2x^2+7x+5\right)\)
\(=-\left(2x^2+2x+5x+5\right)\)
\(=-\left[2x\left(x+1\right)+5\left(x+1\right)\right]\)
\(=-\left(x+1\right)\left(2x+5\right)\)
22 tháng 12 2022
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
9 tháng 9 2023
a: 3x^2-9
=3*x^2-3*3
=3(x^2-3)
b: 1/2x^2-2y^2
=1/2(x^2-4y^2)
=1/2(x-2y)(x+2y)
c: 3x^2-12y^2
=3(x^2-4y^2)
=3(x-2y)(x+2y)
d: 1/3x^2y^2-3x^2
=1/3x^2(y^2-9)
=1/3x^2(y-3)(y+3)
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
GH
27 tháng 7 2023
a Đề sai: )
b
\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)
c
\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)
d
\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)
e
\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)
27 tháng 7 2023
c: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
d: =x^2(x^2+2x+1)
=x^2(x+1)^2
e: =5(x^2-2xy+y^2-z^2)
=5[(x-y)^2-z^2]
=5(x-y-z)(x-y+z)
NH
2
MH
17 tháng 9 2021
a) \(\left(x+3\right)\left(5x-1\right)=\left(5x+1\right)\left(x-2\right)\)
\(\left(x+3\right)\left(5x-1\right)-\left(5x+1\right)\left(x-2\right)=0\)
\(\left(x+3\right)\left(5x-1\right)+\left(5x-1\right)\left(x-2\right)=0\)
\(\left(x+3+x-2\right)\left(5x-1\right)=0\)
\(\left(2x-1\right)\left(5x-1\right)=0\)
Xảy ra 2 trường hợp:
TH1:2x-1=0⇒x=\(\dfrac{1}{2}\)
TH2:5x-1=0⇒x=\(\dfrac{1}{5}\)
17 tháng 9 2021
b: Ta có: \(\left(2x-3\right)\left(x+1\right)=2x\left(x-1\right)\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+2x=0\)
\(\Leftrightarrow x=3\)
MH
25 tháng 10 2021
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
25 tháng 10 2021
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
a, \(5x-15y=5\left(x-3y\right)\)
b, \(12y\left(2x-5y\right)+6xy\left(5-2x\right)=12y\left(2x-5\right)-6xy\left(2x-5\right)\)
\(=6y\left(2-x\right)\left(2x-5\right)\)
c, \(x^2-7x+12=x^2-3x-4x+12=\left(x-4\right)\left(x-3\right)\)