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\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
a: =x^2(x^2+2x+1)
=x^2(x+1)^2
b: =x^3+3x^2y+3xy^2+y^3-x-y
=(x+y)^3-(x+y)
=(x+y)[(x+y)^2-1]
=(x+y)(x+y-1)(x+y+1)
c: =5(x^2-2xy+y^2-4z^2)
=5(x-y-2z)(x-y+2z)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
a) \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
a) \(^{x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)}\)
b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)
c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)
\(=-5\left(4z^2+2xy-y-x^2\right)\)
d)\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3xy^2+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a, \(14x^2y-21xy^2-28x^2y^2=7xy\left(2x-3y-4xy\right)\)
b,\(x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
c,\(10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x-8\right)\)
d,\(\left(3x+1\right)^2-\left(x+1\right)^2=9x^2+6x+1-x^2-2x-1=7x^2-4x=x\left(7x-4\right)\)
e,\(x^3+y^3+z^3-3xyz=\)
x
5x^2 - 10xy + 5y^2 - 20z^2
=5[(x-y)^2 - (2z)^2]=5(x-y-2z)(x-y+2z)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
a) \(x^4+2x^3+x^2=\left(x^2\right)^2+2.x^2.x+x^2=\left(x^2+x\right)^2\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y=x^3+3x^2y+3xy^2+y^3-x-y\)
\(=\left(x-y\right)^3-\left(x+y\right)\)
c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5}x-\sqrt{5}y\right)^2-20z^2\)
Câu b :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
Câu c :
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2\right)-20z^2\)
\(=5\left(x-y\right)^2-20z^2\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
a: \(x^4+2x^3+x^2=x^2\left(x+1\right)^2\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3)
b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
f.2(x^5)-x^2-5x ( mik ko bik làm)
g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)
\(=\left(x^2-2\right)-x^2=\left(x^2-2+x\right)\left(x^2-2-x\right)\)