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XO
24 tháng 5 2021
a) x2 + xy + y - 1 = (x2 - 1) + (xy + y) = (x - 1)(x + 1) + y(x + 1) = (x + 1)(x + y - 1)
b) 4 - x2 + 2xy - y2 = 4 - (x2 - 2xy + y2) = 4 - (x - y)2 = (x - y + 2)(4 - x + y)
c) 8x2 - 18y2 = 2(4x2 - 9y2) = 2[(2x)2 - (3y)2] = 2(2x - 3y)(2x + 3y)
d) 8x3 - 4x2 - 6xy - 9y2 - 27y3
= (8x3 - 27y3) - (4x2 + 6xy + 9y2)
= (2x - 3y)(4x2 + 6xy + 9y2) - (4x2 + 6xy + 9y2)
= (2x - 3y - 1)(4x2 + 6xy + 9y2)
e) 4x2 - x - 3 = 4x2 - 4x + 3x - 3 = 4x(x - 1) + 3(x - 1) = (x - 1)(4x + 3)
f) 4x2 - 8x + 3 = 4x2 - 2x - 6x + 3 = 2x(2x - 1) - 3(2x - 1) = (2x - 3)(2x - 1)
20 tháng 5 2022
Bài 1:
a: \(3x\left(x-a\right)+4a\left(a-x\right)\)
=3x(x-a)-4a(x-a)
=(x-a)(3x-4a)
b: \(x^2\left(y^2+z\right)+y^3+yz\)
\(=x^2\left(y^2+z\right)+y\left(y^2+z\right)\)
\(=\left(x^2+y\right)\left(y^2+z\right)\)
c: \(3x^2\left(x+1\right)-5x\left(x+1\right)^2+4\left(x+1\right)\)
\(=\left(x+1\right)\left[3x^2-5x\left(x+1\right)+4\right]\)
\(=\left(x+1\right)\left(3x^2-5x^2-5x+4\right)\)
\(=\left(x+1\right)\left(-2x^2-5x+4\right)\)
16 tháng 8 2019
Bài 4 :
a) \(x^3+x^2y-xy^2-y^3=x^2\left(x+y\right)-y^2\left(x+y\right)=\left(x^2-y^2\right)\left(x+y\right)=\left(x-y\right)\left(x+y\right)^2\)
b)\(x^2y^2+1-x^2-y^2=\left(x^2y^2-x^2\right)-\left(y^2-1\right)=x^2\left(y^2-1\right)-\left(y^2-1\right)=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)
c) \(x^2-y^2-4x+4y=\left(x^2-y^2\right)-\left(4x-4y\right)=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)=\left(x-y\right)\left(x+y-4\right)\)
d)
\(x^2-y^2-2x-2y=\)\(\left(x^2-y^2\right)-\left(2x+2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
e) Trùng câu d
f) \(x^3-y^3-3x+3y=\left(x-y\right)\left(x^2-xy+y^2\right)-3\left(x-y\right)=\left(x-y\right)\left(x^2-xy+y^2-3\right)\)
16 tháng 8 2019
Bài 5:
a) \(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow x=\frac{9}{2}\)
vậy........
c) \(x^4+2x^3-6x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
Vậy
d) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy ........
21 tháng 7 2022
a: \(=7x\left(xy-3\right)\)
d: \(=\left(x+1\right)\left(10x-8y\right)\)
\(=2\left(5x-4y\right)\left(x+1\right)\)
e: \(=\left(x-100\right)\cdot7x\)
f: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
RM
2 tháng 9 2018
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x^2+2\right)^2-y^4\)
\(=\left(x^2+y^2+2\right)\left(x^2-y^2+2\right)\)
16 tháng 9 2018
\(\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
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hk tôt
27 tháng 10 2018
Ấn nhầm :v
a) \(4x^4-21x^2y^2+y^4\)
\(=\left(2x^2\right)^2-2\cdot2x^2\cdot y^2+y^2-25x^2y^2\)
\(=\left(2x^2-y^2\right)^2-\left(5xy\right)^2\)
\(=\left(2x^2-5xy-y^2\right)\left(2x^2+5xy-y^2\right)\)
b) \(x^5-5x^3+4x\)
\(=x^5-4x^3-x^3+4x\)
\(=x^3\left(x^2-4\right)-x\left(x^2-4\right)\)
\(=\left(x^2-4\right)\left(x^3-x\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x^2-1\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
10 tháng 11 2016
Bài 2:
\(A=-2x^2+3x-5\)
\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)
\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)
Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)
a) x2 - 7xy - 18y2
= x2 + 2xy - 9xy - 18y2
= x(x + 2y) - 9y(x + 2y)
= (x - 9y)(x + 2y)
b) 4x2 + 8x - 5
= 4x2 - 2x + 10x - 5
= 2x(2x - 1) + 5(2x - 1)
= (2x + 5)(2x - 1)
c) 4x4 - 21x2y2 + y4
= (4x4 + 4x2y2 + y4) -25x2y2
= (2x2 + y2) - (5xy)2
= (2x2 + 5xy + y2)(2x2 - 5xy + y2)
= \(2\left(x^2+\frac{5}{2}xy+\frac{y^2}{2}\right)2\left(x^2-\frac{5}{2}xy+\frac{y^2}{2}\right)\)
= \(4\left[\left(x+\frac{5}{4}y\right)^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\left[\left(x-\frac{5}{4}\right)y^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\)
\(=4\left(x+\frac{5}{4}y-\frac{\sqrt{17}}{4}y\right)\left(x+\frac{5}{4}y+\frac{\sqrt{17}}{4}y\right)\left(x-\frac{5}{4}y-\frac{\sqrt{17}y}{4}\right)\left(x-\frac{5}{4}y+\frac{\sqrt{17y}}{4}\right)\)
Trả lời:
a, x2 - 7xy - 18y2
= x2 - 9xy + 2xy - 18y2
= ( x2 - 9xy ) + ( 2xy - 18y2 )
= x ( x - 9y ) + 2y ( x - 9y )
= ( x + 2y ) ( x - 9y )
b, 4x2 + 8x - 5
= 4x2 + 10x - 2x - 5
= ( 4x2 + 10x ) - ( 2x + 5 )
= 2x ( 2x + 5 ) - ( 2x + 5 )
= ( 2x - 1 ) ( 2x + 5 )