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19 tháng 7 2018
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
19 tháng 7 2018
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
14 tháng 1 2023
Sửa đề: Tìm cặp \(x,y\in Z\) thỏa mãn \(x^2+3xy+2y^2+3x+6y-4=0\).
\(x^2+3xy+2y^2+3x+6y-4=0\)
\(\Leftrightarrow x^2+2xy+xy+2y^2+3x+6y=4\)
\(\Leftrightarrow\left(x^2+2xy\right)+\left(xy+2y^2\right)+\left(3x+6y\right)=4\)
\(\Leftrightarrow x\left(x+2y\right)+y\left(x+2y\right)+3\left(x+2y\right)=4\)
\(\Leftrightarrow\left(x+2y\right)\left(x+y+3\right)=4\)
Vì \(x,y\in Z\Rightarrow\left(x+2y\right)\left(x+y+3\right)\in Z\)
Trường hợp 1: \(\left\{{}\begin{matrix}x+2y=1\\x+y+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\) (thỏa mãn)
Trường hợp 2: \(\left\{{}\begin{matrix}x+2y=4\\x+y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=6\end{matrix}\right.\) (thỏa mãn)
Trường hợp 3: \(\left\{{}\begin{matrix}x+2y=2\\x+y+3=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\) (thỏa mãn)
Trường hợp 4: \(\left\{{}\begin{matrix}x+2y=-2\\x+y+3=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=3\end{matrix}\right.\) (thỏa mãn)
Vậy: \(\left(x,y\right)=\left[\left(1;0\right),\left(-8;6\right),\left(-4;3\right),\left(-8;3\right)\right]\)
19 tháng 7 2018
Bài 10 :
Câu a :
\(5xy\left(x-y\right)-2x+2y\)
\(=5xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(5xy-2\right)\)
Câu b :
\(6x-2y-x\left(y-3x\right)\)
\(=2\left(3x-y\right)+x\left(3x-y\right)\)
\(=\left(3x-2y\right)\left(2+x\right)\)
Câu c :
\(x^2+4x-xy-4y\)
\(=x\left(x+4\right)-y\left(x+4\right)\)
\(=\left(x+4\right)\left(x-y\right)\)
Câu d :
\(3xy+2z-6y-xz\)
\(=\left(3xy-6y\right)-\left(xz-2z\right)\)
\(=3y\left(x-2\right)-z\left(x-2\right)\)
\(=\left(x-2\right)\left(3y-z\right)\)
19 tháng 7 2018
Bài 11 :
Câu a :
\(4-9x^2=0\)
\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ........................
Câu b :
\(x^2+x+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy........................
Câu c :
\(2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..................
Câu d :
\(3x\left(x-4\right)-x+4=0\)
\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy................................
Câu e :
\(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy........................
Câu f :
\(\left(3x-y\right)^2-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(3x-y-x+y\right)\left(3x-y+x-y\right)=0\)
\(\Leftrightarrow2x\left(4x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy..........................
18 tháng 10 2021
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
VV
15 tháng 10 2018
Bài 1:
a) x( x - y) + x - y = (x - y)(x + 1)
b) 2x + 2y - x( x + y) = ( 2x + 2y) - x( x + y)
= 2( x + y ) - x( x + y ) = ( x + y )(2 - x )
c) 5x2 - 5xy - 10x + 10y = ( 5x2 - 5xy ) - ( 10x - 10y)
= 5x( x - y ) - 10( x - y ) = ( x - y )(5x - 10 )
= 5( x - y )( x - 2 )
d) 4x2 + 6xy - 3x - 6y = Mình ko làm được!!! bạn chép có sai đề không
VV
15 tháng 10 2018
Bài 2:
x ( 2x - 7) - 4x + 14 = 0
⇒ 2x2 - 7x - 4x + 14 = 0 ⇒ ( 2x2 - 4x ) - ( 7x - 14 ) = 0
⇒ 2x( x - 2 ) - 7(x - 2) = 0
⇒ (x - 2)(2x - 7) = 0
⇒ \(\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x = 2; x = \(\dfrac{7}{2}\)
HE
0
K
2
KV
16 tháng 8 2023
x³ - 3x²y + 3xy² - y³ - z³
= (x³ - 3x²y + 3xy² - y³) - z³
= (x - y)³ - z³
= (x - y - z)[(x - y)² + (x - y)z + z²]
= (x - y - z)(x² - 2xy + y² + xz - yz + z³)
--------------------
x² - y² + 8x + 6y + 7
= (x² + 8x + 16) - (y² - 6y + 9)
= (x + 4)² - (y - 3)²
= (x + 4 - y + 3)(x + 4 + y - 3)
= (x - y + 7)(x + y + 1)
15 tháng 8 2023
a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)
\(=\left(x-y\right)^3-z^3\)
\(=\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]\)
\(=\left(x-y-z\right)\left(x^2-2xy+y^2+xz-yz+z^2\right)\)
b: \(=x^2+8x+16-y^2+6y-9\)
=(x+4)^2-(y-3)^2
=(x+4+y-3)(x+4-y+3)
=(x+y+1)(x-y+7)
HE
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