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8 tháng 10 2017
\(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x-y\right)^2-4\left(4x-y\right)\)
\(=\left(4x-y\right)\left(4x-y-4\right)\)
8 tháng 10 2017
a) \(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x\right)^2-8xy+y^2+4\left(y-4x\right)\)
\(=\left(4x-y\right)^2+4\left(y-4x\right)\)
\(=\left(y-4x\right)^2+4\left(y-4x\right)=\left(y-4x\right)\left(y-4x+4\right)\)
29 tháng 10 2022
Bài 1:
a: \(=\left(8x+12\right)^2-\left(15x-6\right)^2\)
\(=\left(8x+12-15x+6\right)\left(8x+12+15x-6\right)\)
\(=\left(-7x+18\right)\left(23x+6\right)\)
b: \(=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
d: \(=3x\left(x-y\right)-7\left(x-y\right)=\left(x-y\right)\left(3x-7\right)\)
e: \(=2x\left(x+4y\right)+5\left(x+4y\right)=\left(x+4y\right)\left(2x+5\right)\)
NN
17 tháng 11 2017
a) \(a^2x+a^2y-9x-9y\)
\(=\left(a^2x+a^2y\right)-\left(9x+9y\right)\)
\(=a^2\left(x+y\right)-9\left(x+y\right)\)
\(=\left(x+y\right)\left(a^2-9\right)\)
\(=\left(x+y\right)\left(a-3\right)\left(a+3\right)\)
b) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
c) \(x^2\left(x-3\right)+12-4x\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
d) \(4x\left(x-y\right)+6y\left(x-y\right)\)
\(=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
e) \(5\left(x+y\right)-xy-y^2\)
\(=5\left(x+y\right)-\left(xy+y^2\right)\)
\(=5\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(5-y\right)\)
TK
1
11 tháng 3 2017
d)Áp dụng BĐT AM-GM
\(x^2+1\ge2\sqrt{x^2}=2x\)
\(y^2+4\ge2\sqrt{4y^2}=4y\)
\(z^2+9\ge2\sqrt{9z^2}=6z\)
Nhân theo vế ta có:
\(VT=\left(x^2+1\right)\left(y^2+4\right)\left(z^2+9\right)\ge2x\cdot4y\cdot6z=48xyz=VP\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x^2+1=2x\\y^2+4=4y\\z^2+9=6z\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\\\left(z-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
e)Áp dụng BĐT AM-GM ta có:
\(x+1\ge2\sqrt{x}\)
\(y+1\ge2\sqrt{y}\)
\(x+y\ge2\sqrt{xy}\)
Nhân theo vế ta có:
\(VT=\left(x+1\right)\left(y+1\right)\left(x+y\right)\ge2\sqrt{x}\cdot2\sqrt{x}\cdot2\sqrt{xy}=8xy=VP\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x+1=2\sqrt{x}\\y+1=2\sqrt{y}\\x+y=2\sqrt{xy}\left(x+y\ge0\right)\end{matrix}\right.\)\(\Rightarrow x=y=0\)
OP
8 tháng 8 2016
a.
\(x^4-x^3-x+1=x^3\times\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)=\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)^2\left(x^2+x+1\right)\)
b.
\(8xy^3-5xyz-24y^2+15z=8y^2\times\left(xy-3\right)-5z\left(xy-3\right)=\left(xy-3\right)\left(8y^2-5z\right)\)
c.
\(x^2-y^2+2x+1=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)
d.
\(x^2+2xz-y^2+2ty+z^2-t^2=\left(x+z\right)^2-\left(y-t\right)^2=\left(x+z+y-t\right)\left(x+z-y+t\right)\)
e.
\(2x^2-y^2+xy=2x^2+2xy-y^2-xy=2x\times\left(x+y\right)-y\times\left(x+y\right)=\left(2x-y\right)\left(x+y\right)\)
f.
\(y^2-y-12=y^2-3y+4y-12=y\times\left(y-3\right)+4\times\left(y-3\right)=\left(y-3\right)\left(y+4\right)\)
30 tháng 9 2018
\(x^4-x^3-x+1\)
\(=x^3\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
10 tháng 4 2020
a)
\(12xy-4x^2y+8xy^2\\ =4xy\cdot\left(3-x+2y\right)\)
b)
\(4x\cdot\left(x-2y\right)-8y\cdot\left(x-2y\right)\\ =4\cdot\left(x-2y\right)\cdot\left(x-2y\right)\\ =4\cdot\left(x-2y\right)^2\)
c)
\(25x^2\cdot\left(y-1\right)-5x^3\cdot\left(1-y\right)\\ =-25x^2\cdot\left(1-y\right)-5x^3\cdot\left(1-y\right)\\ =\left(1-y\right)\cdot\left(-25x^2-5x^3\right)\\ =5x^2\left(1-y\right)\cdot\left(-5-x\right)\)
d)
\(3x\cdot\left(a-x\right)+4a\cdot\left(a-x\right)\\ =\left(a-x\right)\cdot\left(3x+4a\right)\)
e)
\(x^3-3x^2+2\\ =x^3-x^2-2x^2+2\\ =x^2\cdot\left(x-1\right)-2\left(x^2-1\right)\\ =x^2\cdot\left(x-1\right)-2\cdot\left(x-1\right)\cdot\left(x+1\right)\\ =\left(x-1\right)\left[x^2-2\cdot\left(x+1\right)\right]\\ =\left(x-1\right)\cdot-\left(x^2+2x+1\right)\\ =\left(x-1\right)\cdot-\left(x+1\right)^2\)
`a, 8xy^2-2x^2y`
`= 2xy ( 4y - x)`
`b, x(x-y) -y(y-x)`
`= x(x-y) + y(x-y)`
`= (x-y)(x+y)`
`c, x(x-1) + (1-x)^2`
`= x(x-1)+(x-1)^2`
`= (x-1) (x+x-1)`
`=(x-1)(2x-1)`